Can someone please verify whether my proof is logically correct? 🙂
Let $gcd(x,y)=1$. If $xy$ is a perfect square, then $x$ and $y$ are perfect squares.
Proof:
If $xy$ is a perfect square, let $xy=a^{2}$ for some $a\in \mathbb{Z}$. By the fundamental theorem of arithmetic, $a$ has a unique prime factorization $a = p_{1}p_{2}\cdots p_{k}$. Then $a^{2}=(p_{1}p_{2}\cdots p_{k})^{2} = p_{1}^{2}p_{2}^{2}\cdots p_{k}^{2}$. Since $x$ and $y$ are relatively prime, they do not share the same prime factors (i.e. $x=p_{1}p_{4}\cdots p_{k-1}$ and $y=p_{2}p_{3}\cdots p_{k}$). Then $x$ and $y$ each have prime factorizations with each prime having exponent $2$. Therefore, $x$ and $y$ are perfect squares. $\square$
Corrected proof:
If $xy$ is a perfect square, by the fundamental theorem of arithmetic, $xy$ has a unique prime factorization $xy = p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}$ with each prime being distinct and having even exponent. Since $x$ and $y$ are relatively prime, they do not share the same prime factors. Then $x$ and $y$ each have prime factorizations with each prime having even exponents, such as $x=p_{1}^{s_{1}}p_{2}^{s_{2}}\cdots p_{k}^{s_{k}}$ and $y=p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots p_{k}^{r_{k}}$ with $s_{i}+r_{i}=e_{i}$ for all $i=1,…,k$. Therefore, $x$ and $y$ are perfect squares. $\square$
Best Answer
The statement
is not necessarily true; it should be replaced with
Preceding statements should be edited accordingly, bearing in mind that the prime factors of $x^{1/2}$ might not all be distinct; similarly for $y$.