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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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I'm extremely confident that they want you to use $$ x^4 + 4 y^4 = (x^2 - 2 x y + 2 y^2) (x^2 + 2 x y + 2 y^2).$$ –  Will Jagy Apr 2 '12 at 4:40
    
Looks good. Even though it is pretty obvious, you should probably deal with all combinations of primes dividing two of the variables, not just deal with the variables $x$ and $z$. You have of course pushed the descent argument onto an already proved result, but if it has already been proved in detail in the course, that's fine. But it is not fine if quoting the theorem involves looking ahead. –  André Nicolas Apr 2 '12 at 5:03
    
@Will I would never have seen that and if I had, I'm not sure what I would have done with it from there. Hmm... So $z^2=(x^2+2y^2)^2-(2xy)^2$. Not coming to me. I figured if there was a theorem from the section it wanted me to use, it would be that the area of a Pythagorean triple is never equal to a square number. Wait... Aw crap... I did all this for nothing. This right triangle would have area $(xy)^2$. –  Mike Apr 2 '12 at 5:04
    
@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. –  Mike Apr 2 '12 at 5:10

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