# Proof help involving perfect squares

I'm self-studying abstract algebra (using the free book found on this website: http://abstract.pugetsound.edu/), and am having difficulty solving Problem 19 of Chapter 2's end-of-chapter questions. I have quoted it below:

Let $$x,y \in \mathbb{N}$$ be relatively prime. If $$x\cdot y$$ is a perfect square, prove that $$x$$ and $$y$$ must both be perfect squares.

The answer/hint in the back of the book is "Use the fundamental theorem of arithmetic."

### What I've done:

To me, this fact is obvious. However, I don't know how to represent this in mathematical terms/symbols. In words, I would say:

1. To be a perfect square, the number in question must have a prime factorization consisting only of primes raised to even powers.
2. Suppose $$x$$ is not a perfect square (that is, some prime factor (we will call $$p$$) of $$x$$ is not raised to an even power).
3. To make $$x\cdot y$$ a perfect square, $$p$$ must also exist as a prime factor of $$y$$. This is a contradiction, as $$x$$ and $$y$$ are coprime. The same reasoning applies if $$y$$ is assumed as not a perfect square.

### My question:

First, I'd like to know if this is a valid proof. (To me, it appears solid, but there may be something I'm overlooking.)

Second, I'd like to know if there is a more "mathematical" way of representing this proof. (Symbols, etc. that I should use.)

• It looks just fine to me, and I can't think of some 'More mathematical" proon than one using the FTA. Dec 4, 2012 at 14:59
• Is there a proof that uses Bezout's identity ?
– Amr
Dec 4, 2012 at 15:14

Your math is fine, although you might want to mention that you are invoking the fundamental theorem of arithmetic when you factor $x$, $y$, and $xy$ (uniquely) into primes.
If $x$ is not a square then $x = p_1^{\alpha_1} \dots p_k^{\alpha_k}$, where at least one of $\alpha_1, \dots , \alpha_k$ are odd. Likewise, $y = q_1^{\beta_1} \dots q_m^{\beta_m}$. Note that $p_i \neq q_j$ as $\gcd(x,y) = 1$. Thus, $xy = p_1^{\alpha_1} \dots p_k^{\alpha_k} q_1^{\beta_1} \dots q_m^{\beta_m}$. Since at least one of the $\alpha_i$ were odd, and $p_i \neq q_j$ for all $(i,j)$, it follows that $xy$ is also not a perfect square.