It's probably a good thing I decided to try working on problems from the book. This section seems to be proving difficult. In any case, I'm asked to prove that
has no non-trivial integer solutions. I want to make sure I have this right. I start by trying to prove if non-trivial solutions exist, there exist solutions where $x$, $y$, and $z$ are coprime. My first step is to assume there exists an odd prime $p$ such that $p|z^2$ and $p|x^4$. Then $p|z$, $p|x$, $p|4y^4$, and $p|y$. Then $p^4|y^4$, $p^4|x^4$, and $p^4|z^2$. If $p^4|z^2$, $p^2|z$. So there exist integers $a$, $b$, and $c$ such that
So I can divide out all odd primes. Now if $2|x$, $2|z$. So integers $a$ and $b$ exist such that
Eventually, $z^2$ will run out of factors of 2, so there must be a solution (if any exist) where $x$ and $z$ are odd. Therefore, if solutions exist, there must be a solution where $x$, $y$, and $z$ share no common factors. Then it follows that $(x^2,2y^2,z)$ is a primitive Pythagorean triple. So there exist $m$ and $n$ such that
Now since $y^2=mn$ and $\gcd(m,n)=1$, $m$ and $n$ must be square numbers. So let $m=a^2$ and $n=b^2$. Then $x^2=a^4-b^4$. But there's a theorem proven in the book that states this equation has no integer solutions. Therefore, there are no solutions where $x$, $y$, and $z$ are coprime which in turn means there are no non-trivial solutions.
So again, is this all correct?