$a =\prod_{i = 1}^{r} p_{i}^{a_i}$ with $a_{i} > 0$ for each $i$ is the canonical representation of $a$...

Question

$a =\prod_{i = 1}^{r} p_{i}^{a_i}$ with $a_{i} > 0$ for each $i$ is the canonical representation of $a$,Prove that $a$ is the square of an integer if and only if $a_i$ is even for each $i$.

-The section we are currently studying involved purely numerical GCD and LCM questions along with some questions involving finding the number of positive divisors and sum of positive divisors. we did not although go over proofs such as this and I am a bit lost as to how to structure it. Any help is really appreciated.

Answer 1

Well think about how $p^3$ could be a square. If you see why that can't be a square you'll know how to solve this in general. You see the only non-trivial way to factor $p^3$ is as $p\cdot p^2$ and those terms aren't equal so it's not a square. Now think about how you could factor $p^d$ where $d$ is odd. You can't factor it as $p^{d/2}p^{d/2}$ because $d/2$ is not an integer, but that's what you'd need to do to factor it as a square.

Answer 2

Suppose that $a = \prod p_i^{a_i} = b^2$ for some $b \in \Bbb N$. Write $b = \prod q_i^{b_i}$ (canonically). Then, $\prod q_i ^{2b_i} = \prod p_i ^{a_i}$.

It must be that each $p_i^{a_i}$ is a divisor of some $q_t^{2b_t}$, hence each $q_t$ is a power of some $p_i$ and there are as many $q_i$'s as there are $p_i$'s. By comparison, each $a_i$ is even.