Q:

A question in Elementary Number Theory

I have a question as follow:
Suppose I have two positive integers a, b (both not equal to 1) such that a divides b^2. Show that a and b have common divisors.
It is straightforward to use prime factorisation to prove. However, I am seeking for alternative explanation.
Thank you for your suggestions.

A:

Because of $$a|b^2$$ there is a positive integer $k$ with $$ak=b^2$$ Hence, we have $$\frac{k}{b}=\frac{b}{a}$$
If we assume $gcd(a,b)=1$, we can conclude $a|b$ because every fraction equal to $\frac{b}{a}$ is of the form $\frac{mb}{ma}$, hence $ma=b$ for some $m$. 
Hence $a\ne 1$ is a common divisor of $a$ and $b$. This is a contradiction to $gcd(a,b)=1$, hence $gcd(a,b)>1$ completing the proof.

