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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

$x^4+4y^4=z^2$

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

$x=pa,y=pb,z=p^2c$

$p^4a^4+4p^4b^4=p^4c^2, a^4+4b^4=c^2$.

So I can divide out all odd primes. Now if $2|x$, $2|z$. So integers $a$ and $b$ exist such that

$x=2a,z=2b$

$16a^4+4y^4=4b^2, y^4+4a^4=b^2$.

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

$\gcd(m,n)=1,x^2=m^2-n^2,2y^2=2mn,y^2=mn,z=m^2+n^2$

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?

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    @AndréNicolas Nope, the theorem was from the same section. And proved using the theorem I mentioned from my last comment. The theorem I should have used. The proof given was not at all obvious. It would have been an easier application to this problem. – 2012-04-02

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