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Battle of Alamance
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Battle of Alamance
Part of the War of the Regulation

"Alamance, The First Battle of the Revolution, Burlington, N.C., From the original drawing by J. Steeple Davis", written on a postcard, circa 1905-1915
Date
May 16, 1771
Location
near Great Alamance Creek in Alamance County, North Carolina
Result
Decisive government victory
Belligerents
North Carolina Provincial Militia North Carolina Regulators
Commanders and leaders
Royal Governor William Tryon Herman Husband, Captain Montgomery, others
Strength
appx. 1,000 appx. 2,000
Casualties and losses
Between 9 and 27 killed
61 wounded 9 killed
Unknown number injured
7 later executed for treason
[hide]
v
t
e
War of the Regulation
Yadkin River
Alamance
The site of the Battle of Alamance, including red flags, to the right, marking militia positions and an 1880 commemorative monument, in the distance, to the far left.
The Battle of Alamance was the final battle of the War of the Regulation, a rebellion in colonial North Carolina over issues of taxation and local control. Some historians in the late nineteenth-early twentieth centuries considered the battle to be the opening salvo of the American Revolution,[1] and locals agreed with this assessment.[2] Yet, this has been questioned by present-day historians arguing that the Regulators (though viewed in the eyes of the royal governor and his allies as being in rebellion against King, country, and law) were not intending a complete overthrow of His Majesty's Government in North Carolina. They were only standing up against those certain local officials who had become corrupt and unworthy tools of the King, and they only turned to riot and armed rebellion as a last resort when all other peaceful means through petitions, elections to the Assembly, etc. had failed to redress their grievances. Many surviving ex-Regulators became loyalists during the Revolution, and several anti-Regulators [e.g. William Hooper, Alexander Martin, and Francis Nash] became patriots during the Revolution.[3] Named for nearby Great Alamance Creek, the battle took place in what was then Orange County and has since become Alamance County in the central Piedmont about 6 miles (9.7 km) south of present-day Burlington, North Carolina.
Contents
[hide]
1 Events prior to the battle
2 Course of the battle
3 Participants in the battle
3.1 Colonial Militia
3.2 Regulators
4 Memorial
5 See also
6 References
7 External links
Events prior to the battle[edit]
In the spring of 1771, Royal Governor William Tryon left New Bern, mustering and marching approximately 1,000 militia troops west to address a rebellion that had been brewing in western counties for several years, but which had included only minor, scattered acts of violence, followed by refusals to pay fees, disruptions of court proceedings, and continued harassment of government officials. About 2,000 so-called "regulators" had gathered, hoping to gain concessions from the Governor by intimidating him with a show of superior force. Funded by council member and wealthy merchant Samuel Cornell for £6,000,[4] on May 11, Tryon left the county seat of Hillsborough with his militia to confront the Regulators, who had made camp south of Great Alamance Creek in western Orange County (present-day Alamance County).[5]
Course of the battle[edit]
On the evening of May 15, Tryon received word that the Regulators were camped about six miles away. The next morning, at about 8:00 am, Tryon's troops set out to a field about one-half mile from the camp of the Regulators. He formed two lines, and divided his artillery between the wings and the center of the first line. The Regulators remained disorganized, with no leadership – no officer ranked higher than captain – and no anticipation of an attack, expecting that their superior numbers would frighten Tryon's militia.
Tryon sent one of his aides-de-camp, Captain Philemon Hawkins II, and the Sheriff of Orange County with a proclamation:
Alamance Camp, Thursday, May 16, 1771.
To Those Who Style Themselves "Regulators": In reply to your petition of yesterday, I am to acquaint you that I have ever been attentive to the interests of your County and to every individual residing therein. I lament the fatal necessity to which you have now reduced me by withdrawing yourselves from the mercy of the crown and from the laws of your country. To require you who are now assembled as Regulators, to quietly lay down your arms, to surrender up your leaders, to the laws of your country and rest on the leniency of the Government. By accepting these terms within one hour from the delivery of this dispatch, you will prevent an effusion of blood, as you are at this time in a state of rebellion against your King, your country, and your laws.
(Signed) William Tryon.
While the terms were being read, Tryon's troops began to move forward. Shortly after that, Tryon was informed that the Regulators had rejected his terms. Herman Husband, a Quaker, realizing violence was about to take place, left the area.
By midday the hour had expired. Tryon sent one final warning:
Gentlemen and Regulators: Those of you who are not too far committed should desist and quietly return to your homes, those of you who have laid yourselves liable should submit without resistance. I and others promise to obtain for you the best possible terms. The Governor will grant you nothing. You are unprepared for war! You have no cannon! You have no military training! You have no commanding officers to lead you in battle. You have no ammunition. You will be defeated!
Some of the Regulators petitioned the Royal Governor to give up seven captured Regulators in exchange for two of his men that they had captured the previous day. Tryon agreed, but after a half an hour, the captured officers did not appear. He became suspicious that his positions were being flanked and ordered the militia to march within 30 yards of the Regulators.[6] Shortly thereafter, a large crowd of Regulators appeared in front of the militia, waving their hats and daring the militia to open fire.
At about this time, two men who had been attempting to negotiate a peace between the two sides left Tryon's camp: Reverend Caldwell and Robert Thompson. Caldwell made it to the field between the two lines, but was warned by the Regulators, who saw that the Governor was about to open fire. Thompson was detained by Tryon as a prisoner. Tryon, in a moment of anger, took a musket from a militiaman and shot Thompson dead. Realizing what he had done, he sent a flag bearer named Donald Malcolm with a white flag in hopes of calming things quickly. The flag bearer was himself fired upon by the Regulators, who called out, "Fire and be damned".
The Regulators lacked the leadership, organization, and ammunition that Tryon had, but the early course of the battle went well for them. They employed what was referred to as "Indian style" fighting, hiding behind trees and avoiding structure and lines. This allowed two of the Regulators, brothers named McPherson, to capture one of Tryon's three cannons. Unfortunately for them, the Regulators had no ammunition and it could not be used.[7]
A man considered one of the principal military leaders of the Regulators, Captain Montgomery, was killed by a shell at about the same time a bullet hit Tryon's hat. The Governor sent a second white flag, but the aide-de-camp was killed while regulator Patrick Muller called for his fellow insurgents to cease fire. Outraged at the disregard of a second white flag, the Governor rallied his troops against the insurgents, whose ammunition was running out. Many of the Regulators fled the field. Delays prevented the 300 reinforcements under Captain Benjamin Merrill from arriving in time. Some of the Regulators remained behind to continue firing upon the milita. Tryon then ordered the woods to be set on fire.[7]
Losses for both sides are disputed. Tryon reported nine dead and 61 wounded among the militia. Other historians indicate much greater numbers, between 15 and 27 killed.[8] Both sides counted nine dead among the Regulators and from dozens to over one-hundred wounded.
Tryon took 13 prisoners. One of them, James Few, was executed at the camp, and six were executed later in nearby Hillsborough. Many Regulators traveled on to frontier areas beyond North Carolina. The Royal Governor pardoned others and allowed them to stay on the condition that they pledge an oath of allegiance to the royal government.[7]
The battle took place in what was then Orange County. During the American Revolution a decade later, the same section of Orange County (subdivided into Alamance County in 1849) saw several minor skirmishes, including the infamous Pyle's Hacking Match in 1781. Recent archaeological studies at the site have shown that the area now known as Alamance Battleground was also the site of another skirmish in the revolutionary war and of a civil war era Confederate encampment.[9]
Participants in the battle[edit]
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Colonial Militia[edit]
According to Tryon's journal, the following men served under his command:
Major-generals: John Ashe and Thomas Lloyd
Lieutenant-generals: John Rutherford, Lewis Henry deRosset, John Sampson, Robert Palmer, Benjamin Heron, and Samuel Strudwick
Majors of brigade: Abner Nash and Robert Howe
Colonels: Alexander Osborne, Edmund Fanning, Robert Harris, James Sampson, Samuel Spencer, James Moore, and Maurice Moore
Lieutenant-colonels: John Frohock, Moses Alexander, Alexander Lillington, John Gray, Samuel Benton, and Robert Schaw
Majors: William Bullock, Walter Lindsay, Thomas Lloyd, Martin Fifer, and John Hinton
Alexander Lillington and James Moore were both American patriots at the Battle of Moore's Creek Bridge
Richard Caswell was delegate to the Continental Congress in Philadelphia, one of the principal authors of the 1776 constitution of North Carolina, and the first governor of the newly independent state
Francis Nash, whose guilt for extortion precipitated the War of the Regulation, fought and died as an American patriot in the Revolution
Griffith Rutherford served as a brigadier general in the Continental Army
Regulators[edit]
The following individuals were numbered as members of the Regulators:
Herman Husband
James Hunter – So-called "General of the Regulators", whose 1901 statue is now found at Alamance Battleground
James Few – executed at camp after the battle
Charles Harrington – died from wounds received at the battle
The following were excepted from pardons by Tryon:
Samuel Jones
Joshua Teague
Samuel Waggoner
Simon Dunn Jr.
Abraham Creson
Benjamin Merrit (Merrill)
James Wilkerson Sr.
Edward Smith
John Bumpass
Joseph Boring
William Rankin
William Robeson
John Winkler
John Wilcox
Six men were found guilty of treason, but were pardoned at Tryon's behest:
Forest Mercer
James Stewart
James Emerson – later signed the Revolutionary War Patriots' Muster Roll as James Emison (Emmerson) http://www.ncgenweb.us/chatham/militia.htm
Hermon (or Harmon) Cox – his powder horn is now on display at Alamance Battleground
William Brown
James Copeland
Six men were found guilty of treason and were sentenced to be hanged, drawn and quartered, although in practice, they were only hanged:
Benjamin Merrill
Robert Matear (Matter)
James Pugh
Captain Robert Messer
Peter Craven
One unknown man
Memorial[edit]
Visitors to Alamance Battleground State Historic Site may view the field of battle, memorialized in 1880 with a granite monument and a second monument in 1903. Today the site contains exhibits, period cannon, and colored flags representing troop positions. The visitors' center offers exhibits, artifacts, and a presentation on the battle. Visitors may also tour the onsite Allen House, a restored frontier farmstead of the period.
See also[edit]
Overmountain Men
Watauga Association
References[edit]
Jump up ^ Lutie Andrews McCorkle, "The North Carolina Booklet", Vol III, No. 7, p. 29, 1903
Jump up ^ "First Battle of the Revolution". The Historical Marker Database.
Jump up ^ Sadlier, Sarah (November 2012). "Prelude to the American Revolution? The War of Regulation: A Revolutionary Reaction for Reform" (PDF). The History Teacher. The Society for History Education (SHE). 46 (1): 97–126. Retrieved 17 May 2016.
Jump up ^ "Colonial Williamsburg's latest buy: Money"[permanent dead link], Virginia Gazette, Oct 31, 2010
Jump up ^ "Alamance Battleground: Overview". North Carolina Historic Sites website.
Jump up ^ Patrick O'Kelley, Nothing But Blood and Slaughter: Military Operations and Order of Battle of the Revolutionary War in the Carolinas, Vol 1, 1771–1779, pp. 15–23
^ Jump up to: a b c "Some Neglected History of North Carolina", by William Edward Fitch, first edition, published 1905
Jump up ^ Williamson's History of North Carolina, Vol. 2, p. 150
Jump up ^ "Search of Alamance Battleground yields archaeological jackpot". Times-News. December 3, 2010.
External links[edit]
Alamance Battleground
"An authentick relation of the Battle of Alamance", Virginia Gazette (Williamsburg), June 27, 1771.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Battle_of_Alamance&oldid=808847781"
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1770 in the Thirteen Colonies
1771 in the Thirteen Colonies
Conflicts in 1770
Conflicts in 1771
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text:
Let It Go (Keyshia Cole song)
From Wikipedia, the free encyclopedia
Jump to navigation Jump to search
"Let It Go"
Single by Keyshia Cole featuring Missy Elliott and Lil' Kim
from the album Just Like You
Released
June 19, 2007
Format
Download
CD single
Genre
R&B
hip hop
Length
3:58
Label
Imani
Geffen
Songwriter(s)
Keyshia Cole
Kimberly Jones
Melissa Elliott
Jack Knight
Cainon Lamb
James Mtume
Producer(s)
Missy Elliott
Keyshia Cole singles chronology
"Dreamin'"
(2007) Dreamin'2007 "Let It Go"
(2007) Let It Go2007 "Shoulda Let You Go"
(2007) Shoulda Let You Go2007
Lil' Kim singles chronology
"Whoa"
(2006) Whoa2006 "Let It Go"
(2007) Let It Go2007 "Download"
(2009) Download2009
Missy Elliott singles chronology
"Love Me or Hate Me (Remix)"
(2006) Love Me or Hate Me (Remix)2006 "Let It Go"
(2007) Let It Go2007 "Do It"
(2007) Do It2007
"Let It Go" is a song by American R&B recording artist Keyshia Cole. It was written by Cole, Jack Knight, Cainon Lamb, Lil' Kim, and Missy Elliott for her second album Just Like You (2007) and samples "Juicy Fruit" by Mtume, and "Don't Stop the Music" by Yarbrough and Peoples, while also interpolating "Juicy" by The Notorious B.I.G., who also sampled "Juicy Fruit." Production was handled by Elliott and Lamb, with Lil' Kim and Elliott also appearing as featured vocalists on the track.
Released as the album's lead single, it debuted at number 74 on the Billboard Hot 100 on the issue date of July 7, 2007, and peaked at number 7, becoming Cole's third top ten hit single altogether and first top ten Billboard Hot 100 single as a lead artist. It also became Cole's first number 1 hit on the R&B/Hip-Hop Songs chart. It was nominated for a Grammy Award for Best Rap/Sung Collaboration at the 50th annual ceremony, held in February 2008. "Let It Go" was ranked 59th on Rolling Stone's list of the 100 Best Songs of 2007.[1] The single sold over two million copies in the US and was certified 2x platinum by the Recording Industry Association of America (RIAA).[2]
Contents
1 Background
2 Chart performance
3 Charts
4 Certifications and sales
5 See also
6 References
7 External links
Background[edit]
"Let It Go" is about female empowerment and failed relationships. It is about "letting go" of a relationship if one partner is not there for the other, and is not interested in showing love or respect. The song "Let It Go" was originally intended for Fantasia for her self-titled album, but she had too many songs on that album, and Cole wanted it after Elliott played it for her. According to 50 Cent in a Hot 97 interview the song was also offered to G-Unit artist Olivia who picked another Missy Elliott track instead. Lil' Kim appears on the record to pay homage to Biggie, as the song is similar to his hit "Juicy".[3]
The official "Let It Go" remix—featuring T.I., Missy Elliott and Young Dro—is included on the Just Like You album as the final track. An alternate remix featuring Busta Rhymes and extended verses by Lil' Kim remained unreleased till its Internet leak in 2011.[4]
Chart performance[edit]
The song has become Cole's most successful song to date as a lead artist (her most successful song would be (When You Gonna) Give It Up To Me with Sean Paul), peaking at number 7 on the Billboard Hot 100 and reaching number one on Billboard's Hot R&B/Hip-Hop Songs chart.
Charts[edit]
Chart (2007)
Peak
position
Germany (Official German Charts)[5] 72
US Billboard Hot 100[6] 7
US Mainstream Top 40 (Billboard)[7] 29
US Hot R&B/Hip-Hop Songs (Billboard)[8] 1
Certifications and sales[edit]
Region
Certification
Certified units/Sales
United States (RIAA)[9]
2× Platinum 2,000,000^
*sales figures based on certification alone
^shipments figures based on certification alone
See also[edit]
List of number-one R&B singles of 2007 (U.S.)
References[edit]
Jump up ^ No byline (December 11, 2007). "The 100 Best Songs of 2007" Rolling Stone. Retrieved 2007-12-21
Jump up ^ "Recording Industry Association of America". RIAA. Retrieved 2012-02-25.
Jump up ^ ""The Making of Keyshia Cole's "Let It Go""". Vibe. Archived from the original on 2007-06-28. Retrieved 2007-06-25.
Jump up ^ "Keysha Cole ft Missy Elliott,Lil'kim & Busta Rhymes - Let it Go(Remix)". YouTube. January 10, 2011. Retrieved March 22, 2013.
Jump up ^ "Musicline.de – Keyshia Cole Single-Chartverfolgung" (in German). Media Control Charts. PhonoNet GmbH. Retrieved 2014-10-10.
Jump up ^ "Keyshia Cole Chart History (Hot 100)". Billboard. Retrieved 2014-10-10.
Jump up ^ "Keyshia Cole Chart History (Pop Songs)". Billboard. Retrieved 2014-10-11.
Jump up ^ "Keyshia Cole Chart History (Hot R&B/Hip-Hop Songs)". Billboard. Retrieved 2014-10-10.
Jump up ^ "American album certifications – Lil' Kim – Hard Core". Recording Industry Association of America. Retrieved January 2, 2014. If necessary, click Advanced, then click Format, then select Album, then click SEARCH.
External links[edit]
Lyrics of this song at MetroLyrics
show
v
t
e
Keyshia Cole
Discography
Studio albums
The Way It Is
Just like You
A Different Me
Calling All Hearts
Woman to Woman
Point of No Return
11:11 Reset
Mixtapes
Team Invasion Presents Keyshia Cole
Singles
"Never"
"I Changed My Mind"
"(I Just Want It) To Be Over"
"I Should Have Cheated"
"Love"
"Let It Go"
"Shoulda Let You Go"
"I Remember"
"Heaven Sent"
"Playa Cardz Right"
"You Complete Me"
"Trust"
"I Ain't Thru"
"Take Me Away"
"Enough of No Love"
"Trust and Believe"
"I Choose You"
"Next Time (Won't Give My Heart Away)"
"She"
"You"
Featured singles
"Let's Get Blown"
"Impossible"
"(When You Gonna) Give It Up to Me"
"Last Night"
"Dreamin'"
"I Got a Thang for You"
"Boyfriend/Girlfriend"
"Just Stand Up!"
"Game's Pain"
Promotional songs
"Long Way Down"
Concert tours
I Am Music Tour
A Different Me Tour
Love Letter Tour
Woman to Woman Tour
Related articles
Keyshia Cole: The Way It Is
Keyshia & Daniel: Family First
Daniel Gibson
Book
Category
show
v
t
e
Missy Elliott singles
Discography
Production discography
Awards and nominations
Supa Dupa Fly
"The Rain (Supa Dupa Fly)"
"Sock It 2 Me"
"Beep Me 911"
"Hit Em wit da Hee"
Da Real World
"She's a Bitch"
"All n My Grill"
"Hot Boyz"
Miss E… So Addictive
"Get Ur Freak On"
"Lick Shots"
"One Minute Man"
"Take Away"
"4 My People"
Under Construction
"Work It"
"Gossip Folks"
"Back in the Day"
This Is Not a Test!
"Pass That Dutch"
"I'm Really Hot"
The Cookbook
"Lose Control"
"Teary Eyed"
"We Run This"
Other singles
"Fighting Temptation"
"Ching-a-Ling"
"Shake Your Pom Pom"
"WTF (Where They From)"
"Pep Rally"
"I'm Better"
Featured singles
"That's What Little Girls Are Made Of"
"Brand New"
"Ooh, Ooh Baby"
"The Things That You Do (Bad Boy Remix)"
"Steelo"
"Do Thangz (Main Pass Remix)"
"You Don't Have to Worry (Remix)"
"Cold Rock a Party"
"Can We"
"Not Tonight (Remix)"
"Up Jumps Da Boogie"
"What About Us?"
"Make It Hot"
"5 Minutes"
"Trippin'"
"I Want You Back"
"Here We Come"
"I Like Control"
"Ya Di Ya"
"Take That"
"Is That Yo Chick"
"Son of a Gun"
"Bootylicious (Rockwilder Remix)"
"Oops (Oh My)"
"The Knoc"
"Burnin' Up (Remix)"
"Crew Deep"
"Honk Your Horn"
"Cop That Shit"
"Party to Damascus"
"Tush"
"Turn da Lights Off"
"1, 2 Step"
"Car Wash"
"Touch It (Remix)"
"Love Me or Hate Me (Remix)"
"Let It Go"
"Do It (Remix)"
"Bad Girl"
"Need U Bad"
"Whatcha Think About That"
"Let's Just Do It"
"Work"
"Fakin' It"
"Get Involved"
"Nobody's Perfect"
"Bad Girls (N.A.R.S & Switch Remixes)"
"How Ya Doin'?"
"Without Me"
"Niliria"
"I Deserve It"
"Ghostbusters (I'm Not Afraid)"
Promotional singles
"Best, Best"
"Last Friday Night (T.G.I.F.) (Remix)"
"9th Inning"
"Triple Threat"
"Burnitup!"
show
v
t
e
Lil' Kim
Discography
Awards and nominations
Studio albums and singles
Hard Core
"No Time"
"Crush on You"
"Not Tonight"
The Notorious K.I.M.
"No Matter What They Say"
"How Many Licks?"
La Bella Mafia
"What's The Word"
"The Jump Off"
"Magic Stick"
"Came Back For You"
The Naked Truth
"Lighters Up"/"Shut Up Bitch"
"Whoa"/"Spell Check"
Black Barbie
"Nasty One"
"Go Awff"
Mixtapes
Ms. G.O.A.T.
Black Friday
Hard Core 2K14
Lil Kim Season
Soundtrack and Promotional Singles
"Lady Marmalade"
"In the Air Tonite"
"Download"
"Took Us a Break"
"Spicy"
Features
"Our Time To Shine
"No One Else"
"Float On"
"No More Games"
"It's All About the Benjamins"
"I Can Love You"
"It's All About the Benjamins"
"Money, Power & Respect"
"Hit Em wit da Hee"
"Biggie"
"Notorious B.I.G."
"Get Naked"
"Quiet Storm
"Wait a Minute"
"Get Down on It"
"Kimnotyze"
"Ten Commandments"
"Can't Hold Us Down"
"Do the Damn Ting"
"Sugar (Kimme Sum)"
"Let It Go"
"Stomp"
"Girls"
"Gay-Z"
"Twisted"
"Auction"
"Wake Me Up"
Motion Pictures
She's All That
Zoolander
Those Who Walk in Darkness
Gang of Roses
You Got Served
Carbon Copy
Related articles
Junior M.A.F.I.A.
The Notorious B.I.G.
IRS
Retrieved from "https://en.wikipedia.org/w/index.php?title=Let_It_Go_(Keyshia_Cole_song)&oldid=840776681"
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Lagrange's four-square theorem
From Wikipedia, the free encyclopedia
Jump to: navigation, search
For Lagrange's identity, see Lagrange's identity (disambiguation). For Lagrange's theorem, see Lagrange's theorem (disambiguation).
"four-square theorem" and "four square theorem" redirect here. For other uses, see four square (disambiguation).
Lagrange's four-square theorem, also known as Bachet's conjecture, states that every natural number can be represented as the sum of four integer squares.
p = a 0 2 + a 1 2 + a 2 2 + a 3 2 {\displaystyle p=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}
where the four numbers a 0 , a 1 , a 2 , a 3 {\displaystyle a_{0},a_{1},a_{2},a_{3}} are integers. For illustration, 3, 31 and 310 can be represented as the sum of four squares as follows:
3 = 1 2 + 1 2 + 1 2 + 0 2 31 = 5 2 + 2 2 + 1 2 + 1 2 310 = 17 2 + 4 2 + 2 2 + 1 2 . {\displaystyle {\begin{aligned}3&=1^{2}+1^{2}+1^{2}+0^{2}\\[3pt]31&=5^{2}+2^{2}+1^{2}+1^{2}\\[3pt]310&=17^{2}+4^{2}+2^{2}+1^{2}.\end{aligned}}}
This theorem was proven by Joseph Louis Lagrange in 1770.
Contents
[hide]
1 Historical development
2 The classical proof
3 Proof using the Hurwitz integers
4 Generalizations
5 Algorithms
6 Number of representations
7 Uniqueness
8 Further refinements
9 See also
10 Notes
11 References
12 External links
Historical development[edit]
From examples given in the Arithmetica it is clear that Diophantus was aware of the theorem. This book was translated in 1621 into Latin by Bachet (Claude Gaspard Bachet de Méziriac), who stated the theorem in the notes of his translation. But the theorem was not proved until 1770 by Lagrange.[1]
Adrien-Marie Legendre completed the theorem in 1797–8 with his three-square theorem, by proving that a positive integer can be expressed as the sum of three squares if and only if it is not of the form 4 k ( 8 m + 7 ) {\displaystyle 4^{k}(8m+7)} for integers k {\displaystyle k} and m {\displaystyle m} . Later, in 1834, Carl Gustav Jakob Jacobi discovered a simple formula for the number of representations of an integer as the sum of four squares with his own four-square theorem.
The formula is also linked to Descartes' theorem of four "kissing circles", which involves the sum of the squares of the curvatures of four circles. This is also linked to Apollonian gaskets, which were more recently related to the Ramanujan–Petersson conjecture.[2]
The classical proof[edit]
Several very similar modern versions[3][4][5] of Lagrange's proof exist. The proof below is a slightly simplified version, in which the cases for which m is even or odd do not require separate arguments.
It is sufficient to prove the theorem for every odd prime number p. This immediately follows from Euler's four-square identity (and from the fact that the theorem is true for the numbers 1 and 2).
The residues of a2 modulo p are distinct for every a between 0 and (p − 1)/2 (inclusive). To see this, take some a and define c as a2 mod p. a is a root of the polynomial x2 − c over the field Z/pZ. So is p − a (which is different from a). In a field K, any polynomial of degree n has at most n distinct roots (Lagrange's theorem (number theory)), so there are no other a with this property, in particular not among 0 to (p − 1)/2.
Similarly, for b taking integral values between 0 and (p − 1)/2 (inclusive), the −b2 − 1 are distinct. By the pigeonhole principle, there are a and b in this range, for which a2 and −b2 − 1 are congruent modulo p, that is for which
a 2 + b 2 + 1 2 + 0 2 = n p . {\displaystyle a^{2}+b^{2}+1^{2}+0^{2}=np.}
Now let m be the smallest positive integer such that mp is the sum of four squares, x12 + x22 + x32 + x42 (we have just shown that there is some m (namely n) with this property, so there is a least one m, and it is smaller than p). We show by contradiction that m equals 1: supposing it is not the case, we prove the existence of a positive integer r less than m, for which rp is also the sum of four squares (this is in the spirit of the infinite descent[6] method of Fermat).
For this purpose, we consider for each xi the yi which is in the same residue class modulo m and between (–m + 1)/2 and m/2 (included). It follows that y12 + y22 + y32 + y42 = mr, for some strictly positive integer r less than m.
Finally, another appeal to Euler's four-square identity shows that mpmr = z12 + z22 + z32 + z42. But the fact that each xi is congruent to its corresponding yi implies that all of the zi are divisible by m. Indeed,
z 1 = x 1 y 1 + x 2 y 2 + x 3 y 3 + x 4 y 4 ≡ x 1 2 + x 2 2 + x 3 2 + x 4 2 = m p ≡ 0 ( mod m ) , z 2 = x 1 y 2 − x 2 y 1 + x 3 y 4 − x 4 y 3 ≡ x 1 x 2 − x 2 x 1 + x 3 x 4 − x 4 x 3 = 0 ( mod m ) , z 3 = x 1 y 3 − x 2 y 4 − x 3 y 1 + x 4 y 2 ≡ x 1 x 3 − x 2 x 4 − x 3 x 1 + x 4 x 2 = 0 ( mod m ) , and z 4 = x 1 y 4 + x 2 y 3 − x 3 y 2 − x 4 y 1 ≡ x 1 x 4 + x 2 x 3 − x 3 x 2 − x 4 x 1 = 0 ( mod m ) . {\displaystyle {\begin{alignedat}{4}z_{1}&{}={}x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+x_{4}y_{4}&&{}\equiv {}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}&&{}={}mp\equiv 0&&{\pmod {m}},\\z_{2}&{}={}x_{1}y_{2}-x_{2}y_{1}+x_{3}y_{4}-x_{4}y_{3}&&{}\equiv {}x_{1}x_{2}-x_{2}x_{1}+x_{3}x_{4}-x_{4}x_{3}&&{}={}0&&{\pmod {m}},\\z_{3}&{}={}x_{1}y_{3}-x_{2}y_{4}-x_{3}y_{1}+x_{4}y_{2}&&{}\equiv {}x_{1}x_{3}-x_{2}x_{4}-x_{3}x_{1}+x_{4}x_{2}&&{}={}0&&{\pmod {m}},{\text{ and}}\\z_{4}&{}={}x_{1}y_{4}+x_{2}y_{3}-x_{3}y_{2}-x_{4}y_{1}&&{}\equiv {}x_{1}x_{4}+x_{2}x_{3}-x_{3}x_{2}-x_{4}x_{1}&&{}={}0&&{\pmod {m}}.\end{alignedat}}}
It follows that, for wi = zi/m, w12 + w22 + w32 + w42 = rp, and this is in contradiction with the minimality of m.
In the descent above, we must rule out both the case y1 = y2 = y3 = y4 = m/2 (which would give r = m and no descent), and also the case y1 = y2 = y3 = y4 = 0 (which would give r = 0 rather than strictly positive). For both of those cases, one can check that mp = x12 + x22 + x32 + x42 would be a multiple of m2, contradicting the fact that p is a prime greater than m.
Proof using the Hurwitz integers[edit]
One of the ways to prove the theorem relies on Hurwitz quaternions, which are the analog of integers for quaternions.[7] The Hurwitz quaternions consist of all quaternions with integer components and all quaternions with half-integer components. These two sets can be combined into a single formula
α = 1 2 E 0 ( 1 + i + j + k ) + E 1 i + E 2 j + E 3 k = a 0 + a 1 i + a 2 j + a 3 k {\displaystyle \alpha ={\frac {1}{2}}E_{0}(1+\mathbf {i} +\mathbf {j} +\mathbf {k} )+E_{1}\mathbf {i} +E_{2}\mathbf {j} +E_{3}\mathbf {k} =a_{0}+a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} }
where E 0 , E 1 , E 2 , E 3 {\displaystyle E_{0},E_{1},E_{2},E_{3}} are integers. Thus, the quaternion components a 0 , a 1 , a 2 , a 3 {\displaystyle a_{0},a_{1},a_{2},a_{3}} are either all integers or all half-integers, depending on whether E 0 {\displaystyle E_{0}} is even or odd, respectively. The set of Hurwitz quaternions forms a ring; that is to say, the sum or product of any two Hurwitz quaternions is likewise a Hurwitz quaternion.
The (arithmetic, or field) norm N ( α ) {\displaystyle \mathrm {N} (\alpha )} of a rational quaternion α {\displaystyle \alpha } is the nonnegative rational number
N ( α ) = α α ¯ = a 0 2 + a 1 2 + a 2 2 + a 3 2 {\displaystyle \mathrm {N} (\alpha )=\alpha {\bar {\alpha }}=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}
where α ¯ = a 0 − a 1 i − a 2 j − a 3 k {\displaystyle {\bar {\alpha }}=a_{0}-a_{1}\mathbf {i} -a_{2}\mathbf {j} -a_{3}\mathbf {k} } is the conjugate of α {\displaystyle \alpha } . Note that the norm of a Hurwitz quaternion is always an integer. (If the coefficients are half-integers, then their squares are of the form 1 4 + n : n ∈ Z {\displaystyle {\tfrac {1}{4}}+n:n\in \mathbb {Z} } , and the sum of four such numbers is an integer.)
Since quaternion multiplication is associative, and real numbers commute with other quaternions, the norm of a product of quaternions equals the product of the norms:
N ( α β ) = α β ( α β ¯ ) = α β β ¯ α ¯ = α N ( β ) α ¯ = α α ¯ N ( β ) = N ( α ) N ( β ) . {\displaystyle \mathrm {N} (\alpha \beta )=\alpha \beta ({\overline {\alpha \beta }})=\alpha \beta {\bar {\beta }}{\bar {\alpha }}=\alpha \mathrm {N} (\beta ){\bar {\alpha }}=\alpha {\bar {\alpha }}\mathrm {N} (\beta )=\mathrm {N} (\alpha )\mathrm {N} (\beta ).}
For any α ≠ 0 {\displaystyle \alpha \neq 0} , α − 1 = α ¯ N ( α ) − 1 {\displaystyle \alpha ^{-1}={\bar {\alpha }}\mathrm {N} (\alpha )^{-1}} . It follows easily that α {\displaystyle \alpha } is a unit in the ring of Hurwitz quaternions if and only if N ( α ) = 1 {\displaystyle \mathrm {N} (\alpha )=1} .
The proof of the main theorem begins by reduction to the case of prime numbers. Euler's four-square identity implies that if Langrange's four-square theorem holds for two numbers, it holds for the product of the two numbers. Since any natural number can be factored into powers of primes, it suffices to prove the theorem for prime numbers. It is true for 2 = 1 2 + 1 2 + 0 2 + 0 2 {\displaystyle 2=1^{2}+1^{2}+0^{2}+0^{2}} . To show this for an odd prime integer p {\displaystyle p} , represent it as a quaternion ( p , 0 , 0 , 0 ) {\displaystyle (p,0,0,0)} and assume for now (as we shall show later) that it is not a Hurwitz irreducible; that is, it can be factored into two non-unit Hurwitz quaternions
p = α β . {\displaystyle p=\alpha \beta .}
The norms of p , α , β {\displaystyle p,\alpha ,\beta } are integers such that
N ( p ) = p 2 = N ( α β ) = N ( α ) N ( β ) {\displaystyle \mathrm {N} (p)=p^{2}=\mathrm {N} (\alpha \beta )=\mathrm {N} (\alpha )\mathrm {N} (\beta )}
and N ( α ) , N ( β ) > 1 {\displaystyle \mathrm {N} (\alpha ),\mathrm {N} (\beta )>1} . This shows that both N ( α ) {\displaystyle \mathrm {N} (\alpha )} and N ( β ) {\displaystyle \mathrm {N} (\beta )} are equal to p {\displaystyle p} (since they are integers), and p {\displaystyle p} is the sum of four squares
p = N ( α ) = a 0 2 + a 1 2 + a 2 2 + a 3 2 . {\displaystyle p=\mathrm {N} (\alpha )=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}.}
If it happens that the α {\displaystyle \alpha } chosen has half-integer coefficients, it can be replaced by another Hurwitz quaternion. Choose ω = ( ± 1 ± i ± j ± k ) / 2 {\displaystyle \omega =(\pm 1\pm \mathbf {i} \pm \mathbf {j} \pm \mathbf {k} )/2} in such a way that γ ≡ ω + α {\displaystyle \gamma \equiv \omega +\alpha } has even integer coefficients. Then
p = ( γ ¯ − ω ¯ ) ω ω ¯ ( γ − ω ) = ( γ ¯ ω − 1 ) ( ω ¯ γ − 1 ) . {\displaystyle p=({\bar {\gamma }}-{\bar {\omega }})\omega {\bar {\omega }}(\gamma -\omega )=({\bar {\gamma }}\omega -1)({\bar {\omega }}\gamma -1).}
Since γ {\displaystyle \gamma } has even integer coefficients, ( ω ¯ γ − 1 ) {\displaystyle ({\bar {\omega }}\gamma -1)} will have integer coefficients and can be used instead of the original α {\displaystyle \alpha } to give a representation of p {\displaystyle p} as the sum of four squares.
As for showing that p {\displaystyle p} is not a Hurwitz irreducible, Lagrange proved that any odd prime p {\displaystyle p} divides at least one number of the form u = 1 + l 2 + m 2 {\displaystyle u=1+l^{2}+m^{2}} , where l {\displaystyle l} and m {\displaystyle m} are integers.[7] This can be seen as follows: since p {\displaystyle p} is prime, a 2 ≡ b 2 ( mod p ) {\displaystyle a^{2}\equiv b^{2}{\pmod {p}}} can hold for integers a , b {\displaystyle a,b} , only when a ≡ ± b ( mod p ) {\displaystyle a\equiv \pm b{\pmod {p}}} . Thus, the set X = { 0 2 , 1 2 , … , ( ( p − 1 ) / 2 ) 2 } {\displaystyle X=\{0^{2},1^{2},\dots ,((p-1)/2)^{2}\}} of squares contains ( p + 1 ) / 2 {\displaystyle (p+1)/2} distinct residues modulo p {\displaystyle p} . Likewise, Y = { − ( 1 + x ) : x ∈ X } {\displaystyle Y=\{-(1+x):x\in X\}} contains ( p + 1 ) / 2 {\displaystyle (p+1)/2} residues. Since there are only p {\displaystyle p} residues in total, and | X | + | Y | = p + 1 > p {\displaystyle |X|+|Y|=p+1>p} , the sets X {\displaystyle X} and Y {\displaystyle Y} must intersect.
The number u {\displaystyle u} can be factored in Hurwitz quaternions:
1 + l 2 + m 2 = ( 1 + l i + m j ) ( 1 − l i − m j ) . {\displaystyle 1+l^{2}+m^{2}=(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} ).}
The norm on Hurwitz quaternions satisfies a form of the Euclidean property: for any quaternion α = a 0 + a 1 i + a 2 j + a 3 k {\displaystyle \alpha =a_{0}+a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} } with rational coefficients we can choose a Hurwitz quaternion β = b 0 + b 1 i + b 2 j + b 3 k {\displaystyle \beta =b_{0}+b_{1}\mathbf {i} +b_{2}\mathbf {j} +b_{3}\mathbf {k} } so that N ( α − β ) < 1 {\displaystyle \mathrm {N} (\alpha -\beta )<1} by first choosing b 0 {\displaystyle b_{0}} so that | a 0 − b 0 | ≤ 1 / 4 {\displaystyle |a_{0}-b_{0}|\leq 1/4} and then b 1 , b 2 , b 3 {\displaystyle b_{1},b_{2},b_{3}} so that | a i − b i | ≤ 1 / 2 {\displaystyle |a_{i}-b_{i}|\leq 1/2} for i = 1 , 2 , 3 {\displaystyle i=1,2,3} . Then we obtain
N ( α − β ) = ( a 0 − b 0 ) 2 + ( a 1 − b 1 ) 2 + ( a 2 − b 2 ) 2 + ( a 3 − b 3 ) 2 ≤ ( 1 4 ) 2 + ( 1 2 ) 2 + ( 1 2 ) 2 + ( 1 2 ) 2 = 13 16 < 1. {\displaystyle {\begin{aligned}\mathrm {N} (\alpha -\beta )&=(a_{0}-b_{0})^{2}+(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}+(a_{3}-b_{3})^{2}\\&\leq \left({\frac {1}{4}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}={\frac {13}{16}}<1.\end{aligned}}}
It follows that for any Hurwitz quaternions α , β {\displaystyle \alpha ,\beta } with α ≠ 0 {\displaystyle \alpha \neq 0} , there exists a Hurwitz quaternion γ {\displaystyle \gamma } such that
N ( β − α γ ) < N ( α ) . {\displaystyle \mathrm {N} (\beta -\alpha \gamma )<\mathrm {N} (\alpha ).}
The ring H {\displaystyle H} of Hurwitz quaternions is not commutative, hence it is not an actual Euclidean domain, and it does not have unique factorization in the usual sense. Nevertheless, the property above implies that every right ideal is principal. Thus, there is a Hurwitz quaternion α {\displaystyle \alpha } such that
α H = p H + ( 1 − l i − m j ) H . {\displaystyle \alpha H=pH+(1-l\;\mathbf {i} -m\;\mathbf {j} )H.}
In particular, p = α β {\displaystyle p=\alpha \beta } for some Hurwitz quaternion β {\displaystyle \beta } . If β {\displaystyle \beta } were a unit, 1 − l i − m j {\displaystyle 1-l\;\mathbf {i} -m\;\mathbf {j} } would be a multiple of p {\displaystyle p} , however this is impossible as 1 / p − l / p i − m / p j {\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} } is not a Hurwitz quaternion for p > 2 {\displaystyle p>2} . Similarly, if α {\displaystyle \alpha } were a unit, we would have
( 1 + l i + m j ) H = ( 1 + l i + m j ) p H + ( 1 + l i + m j ) ( 1 − l i − m j ) H ⊆ p H {\displaystyle (1+l\;\mathbf {i} +m\;\mathbf {j} )H=(1+l\;\mathbf {i} +m\;\mathbf {j} )pH+(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} )H\subseteq pH}
so p {\displaystyle p} divides 1 + l i + m j {\displaystyle 1+l\;\mathbf {i} +m\;\mathbf {j} } , which again contradicts the fact that 1 / p − l / p i − m / p j {\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} } is not a Hurwitz quaternion. Thus, p {\displaystyle p} is not Hurwitz irreducible, as claimed.
Generalizations[edit]
Lagrange's four-square theorem is a special case of the Fermat polygonal number theorem and Waring's problem. Another possible generalization is the following problem: Given natural numbers a , b , c , d {\displaystyle a,b,c,d} , can we solve
n = a x 1 2 + b x 2 2 + c x 3 2 + d x 4 2 {\displaystyle n=ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2}}
for all positive integers n {\displaystyle n} in integers x 1 , x 2 , x 3 , x 4 {\displaystyle x_{1},x_{2},x_{3},x_{4}} ? The case a = b = c = d = 1 {\displaystyle a=b=c=d=1} is answered in the positive by Lagrange's four-square theorem. The general solution was given by Ramanujan.[8] He proved that if we assume, without loss of generality, that a ≤ b ≤ c ≤ d {\displaystyle a\leq b\leq c\leq d} then there are exactly 54 possible choices for a , b , c , d {\displaystyle a,b,c,d} such that the problem is solvable in integers x 1 , x 2 , x 3 , x 4 {\displaystyle x_{1},x_{2},x_{3},x_{4}} for all n {\displaystyle n} . (Ramanujan listed a 55th possibility a = 1 , b = 2 , c = 5 , d = 5 {\displaystyle a=1,b=2,c=5,d=5} , but in this case the problem is not solvable if n = 15 {\displaystyle n=15} .[9])
Algorithms[edit]
Michael O. Rabin and Jeffrey Shallit[10] have found randomized polynomial-time algorithms for computing a single representation n = x 1 2 + x 2 2 + x 3 2 + x 4 2 {\displaystyle n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}} for a given integer n {\displaystyle n} , in expected running time O ( ( log  n ) 2 ) {\displaystyle \mathrm {O} ((\log n)^{2})} .
Number of representations[edit]
The number of representations of a natural number n as the sum of four squares is denoted by r4(n). Jacobi's four-square theorem states that this is eight times the sum of the divisors of n if n is odd and 24 times the sum of the odd divisors of n if n is even (see divisor function), i.e.
r 4 ( n ) = { 8 ∑ m ∣ n m if n is odd 24 ∑ m | n m odd m if n is even . {\displaystyle r_{4}(n)={\begin{cases}8\sum \limits _{m\mid n}m&{\text{if }}n{\text{ is odd}}\\[12pt]24\sum \limits _{\begin{smallmatrix}m|n\\m{\text{ odd}}\end{smallmatrix}}m&{\text{if }}n{\text{ is even}}.\end{cases}}}
Equivalently, it is eight times the sum of all its divisors which are not divisible by 4, i.e.
r 4 ( n ) = 8 ∑ m : 4 ∤ m ∣ n m . {\displaystyle r_{4}(n)=8\sum _{m\,:\,4\nmid m\mid n}m.}
We may also write this as
r 4 ( n ) = 8 σ ( n ) − 32 σ ( n / 4 ) , {\displaystyle r_{4}(n)=8\sigma (n)-32\sigma (n/4)\ ,}
where the second term is to be taken as zero if n is not divisible by 4. In particular, for a prime number p we have the explicit formula r4(p) = 8(p + 1).[11]
Some values of r4(n) occur infinitely often as r4(n) = r4(2mn) whenever n is even. The values of r4(n)/n can be arbitrarily large: indeed, r4(n)/n is infinitely often larger than 8 √ log n.[11]
Uniqueness[edit]
The sequence of positive integers which have only one representation as a sum of four squares (up to order) is:
1, 2, 3, 5, 6, 7, 8, 11, 14, 15, 23, 24, 32, 56, 96, 128, 224, 384, 512, 896 ... (sequence A006431 in the OEIS).
These integers consist of the seven odd numbers 1, 3, 5, 7, 11, 15, 23 and all numbers of the form 2 ( 4 k ) , 6 ( 4 k ) {\displaystyle 2(4^{k}),6(4^{k})} or 14 ( 4 k ) {\displaystyle 14(4^{k})} .
The sequence of positive integers which cannot be represented as a sum of four non-zero squares is:
1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41, 56, 96, 128, 224, 384, 512, 896 ... (sequence A000534 in the OEIS).
These integers consist of the eight odd numbers 1, 3, 5, 9, 11, 17, 29, 41 and all numbers of the form 2 ( 4 k ) , 6 ( 4 k ) {\displaystyle 2(4^{k}),6(4^{k})} or 14 ( 4 k ) {\displaystyle 14(4^{k})} .
Further refinements[edit]
Lagrange's four-square theorem can be refined in various ways. For example, Zhi-Wei Sun [12] proved that each natural number can be written as the sum of a sixth power (or a fourth power) and three squares, and Sun's 1-3-5 conjecture (with $1350 prize for the solution) states that any natural number can be written as x 2 + y 2 + z 2 + w 2 {\displaystyle x^{2}+y^{2}+z^{2}+w^{2}} with x , y , z , w {\displaystyle x,y,z,w} nonnegative integers such that x + 3 y + 5 z {\displaystyle x+3y+5z} is a square.
See also[edit]
Fermat's theorem on sums of two squares
Legendre's three-square theorem
15 and 290 theorems
Notes[edit]
Jump up ^ Ireland & Rosen 1990.
Jump up ^ Sarnak 2013.
Jump up ^ Landau 1958, Theorems 166 to 169.
Jump up ^ Hardy & Wright 2008, Theorem 369.
Jump up ^ Niven & Zuckerman 1960, paragraph 5.7.
Jump up ^ Here the argument is a direct proof by contradiction. With the initial assumption that m>2, m<p, is some integer such that mp is the sum of four squares (not necessarily the smallest), the argument could be modified to become an infinite descent argument in the spirit of Fermat.
^ Jump up to: a b Stillwell 2003, pp. 138–157.
Jump up ^ Ramanujan 1917.
Jump up ^ Oh 2000.
Jump up ^ Rabin & Shallit 1986.
^ Jump up to: a b Williams 2011, p. 119.
Jump up ^ Z.-W. Sun 2017.
References[edit]
Hardy, G. H.; Wright, E. M. (2008) [1938]. Heath-Brown, D. R.; Silverman, J. H.; Wiles, Andrew, eds. An Introduction to the Theory of Numbers (6th ed.). Oxford University Press. ISBN 978-0-19-921985-8.
Ireland, Kenneth; Rosen, Michael (1990). A Classical Introduction to Modern Number Theory (2nd ed.). Springer. ISBN 978-1-4419-3094-1. doi:10.1007/978-1-4757-2103-4.
Landau, Edmund (1958) [1927]. Elementary Number Theory. 125. Translated by Goodman, Jacob E. (2nd ed.). AMS Chelsea Publishing.
Niven, Ivan; Zuckerman, Herbert S. (1960). An introduction to the theory of numbers. Wiley.
Oh, Byeong-Kweon (2000). "Representations of Binary Forms by Quinary Quadratic Forms" (PDF). Trends in Mathematics. Information Center for Mathematical Sciences. 3 (1): 102–107.
Rabin, M. O.; Shallit, J. O. (1986). "Randomized Algorithms in Number Theory". Communications on Pure and Applied Mathematics. 39 (S1): S239–S256. doi:10.1002/cpa.3160390713.
Ramanujan, S. (1917). "On the expression of a number in the form ax2 + by2 + cz2 + dw2". Proc. Cambridge Phil. Soc. 19: 11–21.
Sarnak, Peter (2013). "The Ramanujan Conjecture and some Diophantine Equations" (Lecture at Tata Institute of Fundamental Research). ICTS Lecture Series. Bangalore, India.
Stillwell, John (2003). Elements of Number Theory. Springer. ISBN 0-387-95587-9. Zbl 1112.11002. doi:10.1007/978-0-387-21735-2.
Sun, Z.-W. (2017). "Refining Lagrange's four-square theorem". J. Number Theory. 175: 167–190.
Williams, Kenneth S. (2011). Number theory in the spirit of Liouville. London Mathematical Society Student Texts. 76. Cambridge University Press. ISBN 978-0-521-17562-3. Zbl 1227.11002.
External links[edit]
Proof at PlanetMath.org
Another proof
an applet decomposing numbers as sums of four squares
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List of current Indian chief ministers
From Wikipedia, the free encyclopedia
Jump to: navigation, search
Indian states according to party of their chief minister
Bharatiya Janata Party
Communist Party of India (Marxist)
Indian National Congress
Other parties
In the Republic of India, a chief minister is the head of government of each of twenty-nine states and two union territories (Delhi and Puducherry). According to the Constitution of India, at the state-level, the governor is de jure head, but de facto executive authority rests with the chief minister. Following elections to the state legislative assembly, the governor usually invites the party (or coalition) with a majority of seats to form the government. The governor appoints the chief minister, whose council of ministers are collectively responsible to the assembly. Given he has the assembly's confidence, the chief minister's term is usually for a maximum of five years; there are no limits to the number of terms he/she can serve.[1]
Of the thirty-one incumbents, three are women—Mamata Banerjee in West Bengal, Mehbooba Mufti in Jammu and Kashmir, and Vasundhara Raje in Rajasthan. Serving since December 1994 (for 7003831700000000000♠22 years, 281 days), Sikkim's Pawan Kumar Chamling has the longest incumbency. Virbhadra Singh (b. 1934) of Himachal Pradesh is the oldest chief minister,[2] while Arunachal Pradesh's Pema Khandu (b. 1979) is the youngest.[3] Thirteen incumbents belong to the Bharatiya Janata Party, six to the Indian National Congress and two to the Communist Party of India (Marxist); no other party has more than one chief minister in office.
Current Indian chief ministers[edit]
Colour key for parties
Aam Aadmi Party
All India Anna Dravida Munnetra Kazhagam
All India Trinamool Congress
Bharatiya Janata Party
Biju Janata Dal
Communist Party of India (Marxist)
Indian National Congress
Jammu and Kashmir Peoples Democratic Party
Janata Dal (United)
Naga People's Front
Sikkim Democratic Front
Telangana Rashtra Samithi
Telugu Desam Party
State
(past chief ministers)
Name[4]
Portrait
Took office
(tenure length)
Party[a]
Ref
Andhra Pradesh
(list) N. Chandrababu Naidu 000000002014-06-08-00008 June 2014
(7003119900000000000♠3 years, 103 days) Telugu Desam Party [5]
Arunachal Pradesh
(list) Pema Khandu 000000002016-07-17-000017 July 2016
(7002429000000000000♠1 year, 64 days) Bharatiya Janata Party [6][7]
Assam
(list) Sarbananda Sonowal 000000002016-05-24-000024 May 2016
(7002483000000000000♠1 year, 118 days) Bharatiya Janata Party [8]
Bihar
(list) Nitish Kumar 000000002015-02-22-000022 February 2015
(7002940000000000000♠2 years, 209 days) Janata Dal (United) [9]
Chhattisgarh
(list) Raman Singh 000000002003-12-07-00007 December 2003
(7003503500000000000♠13 years, 286 days) Bharatiya Janata Party [10]
Delhi[b]
(list) Arvind Kejriwal 000000002015-02-14-000014 February 2015
(7002948000000000000♠2 years, 217 days) Aam Aadmi Party [11]
Goa
(list) Manohar Parrikar 000000002017-03-14-000014 March 2017
(7002189000000000000♠189 days) Bharatiya Janata Party [12]
Gujarat
(list) Vijay Rupani 000000002016-08-07-00007 August 2016
(7002408000000000000♠1 year, 43 days) Bharatiya Janata Party [13]
Haryana
(list) Manohar Lal Khattar 000000002014-10-26-000026 October 2014
(7003105900000000000♠2 years, 328 days) Bharatiya Janata Party [14]
Himachal Pradesh
(list) Virbhadra Singh 000000002012-12-25-000025 December 2012
(7003172900000000000♠4 years, 268 days) Indian National Congress [15]
Jammu and Kashmir
(list) Mehbooba Mufti 000000002016-04-04-00004 April 2016
(7002533000000000000♠1 year, 168 days) Jammu and Kashmir Peoples Democratic Party [16]
Jharkhand
(list) Raghubar Das 000000002014-12-28-000028 December 2014
(7002996000000000000♠2 years, 265 days) Bharatiya Janata Party [17]
Karnataka
(list) Siddaramaiah 000000002013-05-13-000013 May 2013
(7003159000000000000♠4 years, 129 days) Indian National Congress [18]
Kerala
(list) Pinarayi Vijayan 000000002016-05-25-000025 May 2016
(7002482000000000000♠1 year, 117 days) Communist Party of India (Marxist) [19]
Madhya Pradesh
(list) Shivraj Singh Chouhan 000000002005-11-29-000029 November 2005
(7003431200000000000♠11 years, 294 days) Bharatiya Janata Party [20]
Maharashtra
(list) Devendra Fadnavis 000000002014-10-31-000031 October 2014
(7003105400000000000♠2 years, 323 days) Bharatiya Janata Party [21]
Manipur
(list) N. Biren Singh 000000002017-03-15-000015 March 2017
(7002188000000000000♠188 days) Bharatiya Janata Party [22]
Meghalaya
(list) Mukul Sangma 000000002010-04-20-000020 April 2010
(7003270900000000000♠7 years, 152 days) Indian National Congress [23]
Mizoram
(list) Lal Thanhawla 000000002008-12-07-00007 December 2008
(7003320800000000000♠8 years, 286 days) Indian National Congress [24]
Nagaland
(list) T. R. Zeliang 000000002017-07-19-000019 July 2017
(7001620000000000000♠62 days) Naga People's Front [25]
Odisha
(list) Naveen Patnaik 000000002000-03-05-00005 March 2000
(7003640700000000000♠17 years, 198 days) Biju Janata Dal [26]
Puducherry[b]
(list) V. Narayanasamy 000000002016-06-06-00006 June 2016
(7002470000000000000♠1 year, 105 days) Indian National Congress [27]
Punjab
(list) Amarinder Singh 000000002017-03-16-000016 March 2017
(7002187000000000000♠187 days) Indian National Congress [28]
Rajasthan
(list) Vasundhara Raje 000000002013-12-13-000013 December 2013
(7003137600000000000♠3 years, 280 days) Bharatiya Janata Party [29]
Sikkim
(list) Pawan Kumar Chamling 000000001994-12-12-000012 December 1994
(7003831700000000000♠22 years, 281 days) Sikkim Democratic Front [30]
Tamil Nadu
(list) Edappadi K. Palaniswami 000000002017-02-16-000016 February 2017
(7002215000000000000♠215 days) All India Anna Dravida Munnetra Kazhagam [31]
Telangana
(list) K. Chandrashekar Rao 000000002014-06-02-00002 June 2014
(7003120500000000000♠3 years, 109 days) Telangana Rashtra Samithi [32]
Tripura
(list) Manik Sarkar 000000001998-03-11-000011 March 1998
(7003713200000000000♠19 years, 192 days) Communist Party of India (Marxist) [33]
Uttar Pradesh
(list) Yogi Adityanath 000000002017-03-19-000019 March 2017
(7002184000000000000♠184 days) Bharatiya Janata Party [34]
Uttarakhand
(list) Trivendra Singh Rawat 000000002017-03-18-000018 March 2017
(7002185000000000000♠185 days) Bharatiya Janata Party [35]
West Bengal
(list) Mamata Banerjee 000000002011-05-20-000020 May 2011
(7003231400000000000♠6 years, 122 days) All India Trinamool Congress [36]
Jump up ^ This column names only the chief minister's party. The ministry he heads may be a complex coalition of several parties and independents; those are not listed here.
^ Jump up to: a b Although Delhi and Puducherry each have an elected legislature and a council of ministers (headed by the chief minister), they are officially union territories.
See also[edit]
List of current Indian governors
List of current Indian chief justices
List of current Indian legislative speakers
List of current Indian leaders of the Opposition
List of female Indian chief ministers
List of chief ministers from the Bharatiya Janata Party
List of chief ministers from Communist Party of India (Marxist)
References[edit]
Jump up ^ Durga Das Basu. Introduction to the Constitution of India. 1960. 20th edition, 2011 reprint. pp. 241, 245. LexisNexis Butterworths Wadhwa Nagpur. ISBN 978-81-8038-559-9.
Jump up ^ Ajoy Ashirwad Mahaprashasta. "Bucking the trend". Frontline. Volume 29, issue 6, 24 March – 6 April 2012.
Jump up ^ "Meet Pema Khandu: India’s youngest Chief Minister". The Hindu. 17 July 2016.
Jump up ^ Chief Ministers. India.gov.in. Retrieved on 3 September 2015.
Jump up ^ "Naidu takes oath as Andhra Pradesh CM". The Hindu. 8 June 2014.
Jump up ^ "Pema Khandu sworn in as Chief Minister of Arunachal Pradesh". The Hindu. 17 July 2016.
Jump up ^ "BJP forms govt in Arunachal Pradesh". The Hindu. 31 December 2016.
Jump up ^ "Sarbananda Sonowal sworn in as first BJP CM of Assam". The Hindu. 24 May 2016.
Jump up ^ Kumar, Arun (27 July 2017). "Grand Alliance to NDA: Nitish Kumar changes partner, continues as Bihar CM". Hindustan Times. Patna. Retrieved 27 July 2017.
Jump up ^ Aarti Dhar. "Raman Singh takes oath". The Hindu. 8 December 2003.
Jump up ^ Smriti Kak Ramachandran, Shubhomoy Sikdar. "Kejriwal promises to make Delhi graft-free in 5 years". The Hindu. 14 February 2015.
Jump up ^ Nistula Hebbar, Prakash Kamat. "Parrikar takes oath in Goa as SC declines Cong. plea". The Hindu. 14 March 2017.
Jump up ^ Mahesh Langa. "Vijay Rupani sworn in; Gujarat Cabinet bears Shah’s stamp". The Hindu. 7 August 2016.
Jump up ^ Sarabjit Pandher. "Khattar sworn in". The Hindu. 26 October 2014.
Jump up ^ "Virbhadra Singh sworn in as CM". The Hindu. 26 December 2012.
Jump up ^ "Mehbooba Mufti sworn in as J&K's first woman CM". The Hindu. 4 April 2016.
Jump up ^ Amarnath Tewary. "Raghuvar Das assumes office as CM". The Hindu. 28 December 2014.
Jump up ^ "Siddaramaiah sworn in as Chief Minister of Karnataka". The Hindu. 13 May 2013.
Jump up ^ C. Gouridasan Nair. "Pinarayi takes charge as Kerala Chief Minister". The Hindu. 25 May 2016.
Jump up ^ "Shivraj Chauhan sworn in Chief Minister". The Hindu. 30 November 2005.
Jump up ^ Priyanka Kakodkar. "Uddhav attends Fadnavis swearing-in". The Hindu. 31 October 2014.
Jump up ^ Isha Gupta. "BJP leader Biren Singh sworn in as Manipur Chief Minister". India Today. 15 March 2017.
Jump up ^ "Mukul Sangma sworn in as Chief Minister". The Hindu. 21 April 2010.
Jump up ^ "Lal Thanhawla sworn in as Mizoram chief minister". The Times of India. 11 December 2008.
Jump up ^ "Shurhozelie Liezietsu sworn in as Nagaland Chief Minister". The Hindu. 22 February 2017.
Jump up ^ N. Ramdas. "Naveen Govt. installed". The Hindu. 6 March 2000.
Jump up ^ "Puducherry: V Narayanasamy sworn in as Chief Minister". The Indian Express. 6 June 2016.
Jump up ^ "Amarinder Singh sworn in as Punjab CM". The Hindu. 17 March 2017.
Jump up ^ "Vasundhara swearing-in, a show of strength". The Hindu. 13 December 2013.
Jump up ^ "Pawan Chamling to be longest serving Chief Minister". The Hindu. 17 May 2014.
Jump up ^ T. Ramakrishnan. "Edappadi Palaniswami sworn in as Tamil Nadu Chief Minister". The Hindu. 17 February 2017.
Jump up ^ K. Srinivas Reddy. "KCR sworn in; heads cabinet of 11 ministers". The Hindu. 2 June 2014.
Jump up ^ "Manik Sarkar sworn in as Tripura CM". Rediff.com. 11 March 1998.
Jump up ^ "Yogi Adityanath takes oath as Uttar Pradesh Chief Minister". The Hindu. 19 March 2017.
Jump up ^ Kavita Upadhyay. "Trivendra Singh Rawat takes oath as Uttarakhand Chief Minister". The Hindu. 18 March 2017.
Jump up ^ "Mamata, 37 Ministers sworn in". The Hindu. 21 May 2011.
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Lists of chief ministers of Indian states (and current incumbents)
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text:
Should've Been a Cowboy
From Wikipedia, the free encyclopedia
Jump to navigation Jump to search
"Should've Been a Cowboy"
Single by Toby Keith
from the album Toby Keith
B-side
"Some Kinda Good Kinda Hold on Me"
Released
February 8, 1993
Format
7"
CD
cassette
Recorded
1992
Genre
Country
Length
3:30
Label
Polygram/Mercury 864990
Songwriter(s)
Toby Keith
Producer(s)
Nelson Larkin
Harold Shedd
Toby Keith singles chronology
"Should've Been a Cowboy"
(1993) "He Ain't Worth Missing"
(1993)
"Should've Been a Cowboy"
(1993) "He Ain't Worth Missing"
(1993)
"Should've Been a Cowboy" is a song written and recorded by American country music artist Toby Keith. It was released on February 8, 1993 as his debut single and the first from his self-titled debut album. On June 5, 1993, the song reached number one on the US Billboard Hot Country Songs and the Canadian RPM Country Tracks charts. It also peaked at number 93 on the Billboard Hot 100, making it a minor crossover hit.
It has received more than three million spins on country radio, making it the most-played country song of the 1990s.[1]
The song was formerly a staple following sporting events at Oklahoma State University (home of the Cowboys), often played over the venue's PA system several times in succession as fans emptied the stadium or arena.[2]
Contents
1 Content
2 Music video
3 Other media
4 Chart positions
4.1 Year-end charts
5 Certifications
6 References
7 External links
Content[edit]
The lyrics romanticize the cowboy lifestyle through references to old Westerns. The first verse references Gunsmoke, in which cowboy hero Marshall Dillon never settled down with love interest Miss Kitty. The second verse tells of his own adventures if the narrator were a cowboy: having "a sidekick with a funny name" (possibly a reference to sidekicks like Tonto or Festus or Gordito), travelling west to California (including the historical quote "Go West, young man"), hunting down Western outlaw Jesse James, joining up with the Texas Rangers, and so on. The chorus refers to famous singing cowboys Gene Autry and Roy Rogers, and lists common patterns depicted in Westerns such as six shooters (revolvers) and cattle drives.
Music video[edit]
"Should've Been a Cowboy" was Keith's first music video. It was filmed in Las Cruces, New Mexico and was directed by Marc Ball and premiered on CMT in March 1993. A second version of the video was made featuring clips of the 1992 Dallas Cowboys.[citation needed]
Other media[edit]
The song is available as downloadable content on the music video game, Rock Band.
Chart positions[edit]
"Should've Been a Cowboy" debuted at number 69 on the Hot Country Songs chart for the week of March 6, 1993. It became Keith's first number one single on the chart dated June 5, 1993, remaining there for two weeks. It was certified Gold by the RIAA on November 28, 2012, almost twenty years after its release.[3]
Chart (1993)
Peak
position
Canada Country Tracks (RPM)[4] 1
US Billboard Hot 100[5] 93
US Hot Country Songs (Billboard)[6] 1
Year-end charts[edit]
Chart (1993)
Position
Canada Country Tracks (RPM)[7] 73
US Country Songs (Billboard)[8] 39
Certifications[edit]
Region
Certification
Certified units/Sales
United States (RIAA)[9]
Gold 500,000^
*sales figures based on certification alone
^shipments figures based on certification alone
References[edit]
Jump up ^ Shelburne, Craig (1 March 2007). "Toby Keith Jams, Notches 50 Million Airplays". CMT. Retrieved 4 October 2010.
Jump up ^ McDonnell, Brandy (10 October 2009). "Toby Keith discusses his "weird connection" to OU and OSU". newsok.com. Archived from the original on 10 December 2012. Retrieved 4 June 2012.
Jump up ^ "American single certifications – Toby Keith – Should've Been a Cowboy". Recording Industry Association of America. If necessary, click Advanced, then click Format, then select Single, then click SEARCH
Jump up ^ "Top RPM Country Tracks: Issue 0991." RPM. Library and Archives Canada. June 26, 1993. Retrieved August 5, 2013.
Jump up ^ "Toby Keith Chart History (Hot 100)". Billboard.
Jump up ^ "Toby Keith Chart History (Hot Country Songs)". Billboard.
Jump up ^ "RPM Top 100 Country Tracks of 1993". RPM. December 18, 1993. Retrieved August 5, 2013.
Jump up ^ "Best of 1993: Country Songs". Billboard. Prometheus Global Media. 1993. Retrieved August 5, 2013.
Jump up ^ "American single certifications – Toby Keith – Should%27ve Been a Cowboy". Recording Industry Association of America. Retrieved December 12, 2012. If necessary, click Advanced, then click Format, then select Single, then click SEARCH
External links[edit]
Lyrics of this song at MetroLyrics
hide
v
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e
Toby Keith
Studio albums
Toby Keith
Boomtown
Blue Moon
Dream Walkin'
How Do You Like Me Now?!
Pull My Chain
Unleashed
Shock'n Y'all
Honkytonk University
White Trash with Money
Big Dog Daddy
That Don't Make Me a Bad Guy
American Ride
Bullets in the Gun
Clancy's Tavern
Hope on the Rocks
Drinks After Work
35 MPH Town
Compilation albums
Greatest Hits Volume One
20th Century Masters – The Millennium Collection: The Best of Toby Keith
Greatest Hits 2
35 Biggest Hits
The Bus Songs
Christmas albums
Christmas to Christmas
A Classic Christmas
Singles
"Should've Been a Cowboy"
"He Ain't Worth Missing"
"A Little Less Talk and a Lot More Action"
"Wish I Didn't Know Now"
"Who's That Man"
"Upstairs Downtown"
"You Ain't Much Fun"
"Big Ol' Truck"
"Does That Blue Moon Ever Shine on You"
"A Woman's Touch"
"Me Too"
"We Were in Love"
"I'm So Happy I Can't Stop Crying" (with Sting)
"Dream Walkin'"
"Double Wide Paradise"
"Getcha Some"
"How Do You Like Me Now?!"
"Country Comes to Town"
"You Shouldn't Kiss Me Like This"
"I'm Just Talkin' About Tonight"
"I Wanna Talk About Me"
"My List"
"Courtesy of the Red, White and Blue (The Angry American)"
"Who's Your Daddy?"
"Rock You Baby"
"Beer for My Horses"
"I Love This Bar"
"American Soldier"
"Whiskey Girl"
"Stays in Mexico"
"Mockingbird"
"Honkytonk U"
"As Good as I Once Was"
"Big Blue Note"
"Get Drunk and Be Somebody"
"A Little Too Late"
"Crash Here Tonight"
"High Maintenance Woman"
"Love Me If You Can"
"Get My Drink On"
"She's a Hottie"
"She Never Cried in Front of Me"
"God Love Her"
"Lost You Anyway"
"American Ride"
"Cryin' for Me (Wayman's Song)"
"Every Dog Has Its Day"
"Trailerhood"
"Bullets in the Gun"
"Somewhere Else"
"Made in America"
"Red Solo Cup"
"Beers Ago"
"I Like Girls That Drink Beer"
"Hope on the Rocks"
"Drinks After Work"
"Drunk Americans"
"35 MPH Town"
Related articles
Discography
Show Dog-Universal Music
Krystal Keith
Scotty Emerick
Toby Keith's I Love This Bar & Grill
Broken Bridges (soundtrack)
"Slave to the Habit"
Retrieved from "https://en.wikipedia.org/w/index.php?title=Should%27ve_Been_a_Cowboy&oldid=851214056"
Categories:
1993 singles
Toby Keith songs
Debut singles
Billboard Hot Country Songs number-one singles
RPM Country Tracks number-one singles
Songs written by Toby Keith
Song recordings produced by Harold Shedd
Mercury Records singles
1993 songs
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