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$x, y, z \in \mathbb{N}$,

$\gcd(x, y) = 1$

prove that $x^4 + y^4 = z^2$ has no solutions.

It is true even without $\gcd(x, y) = 1$, but it is easy to see that $\gcd(x, y)$ must be $1$

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    $\begingroup$ Suppose that $x=ku$ and $y=kv$ for some integer $k$. Then $x^4+y^4=k^4(u^4+v^4)$, which is a square if and only if $u^4+v^4$ is a square. $\endgroup$ Jan 12 '13 at 20:03
  • $\begingroup$ @Brian: That looks like an answer to me... $\endgroup$
    – TMM
    Jan 12 '13 at 20:04
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    $\begingroup$ I said the part I cant solve is when GCD = 1. $\endgroup$ Jan 12 '13 at 20:05
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    $\begingroup$ There is a proof here. $\endgroup$
    – P..
    Jan 12 '13 at 20:25
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    $\begingroup$ post an answer. $\endgroup$ Jan 12 '13 at 20:36
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This has been completely revised to match the intended question. The proof is by showing that there is no minimal positive solution, i.e., by infinite descent. It’s from some old notes; I’ve no idea where I cribbed it from in the first place.

Suppose that $x^4+y^4=z^2$, where $z$ is the smallest positive integer for which there is a solution in positive integers. Then $(x^2,y^2,z)$ is a primitive Pythagorean triple, so there are relatively prime integers $m,n$ with $m>n$ such that $x^2=m^2-n^2,y^2=2mn$, and $z=m^2+n^2$.

Since $2mn=y^2$, one of $m$ and $n$ is an odd square, and the other is twice a square. In particular, one is odd, and one is even. Now $x^2+n^2=m^2$, and $\gcd(x,n)=1$ (since $m$ and $n$ are relatively prime), so $(x,n,m)$ is a primitive Pythagorean triple, and it must be $n$ that is even: there must be integers $a$ and $b$ such that $a>b$, $a$ and $b$ are relatively prime, $x=a^2-b^2$, $n=2ab$, and $m=a^2+b^2$. It must be $m$ that is the odd square, so there are integers $r$ and $s$ such that $m=r^2$ and $n=2s^2$.

Now $2s^2=n=2ab$, so $s^2=ab$, and we must have $a=c^2$ and $b=d^2$ for some integers $c$ and $d$, since $\gcd(a,b)=1$. The equation $m=a^2+b^2$ can then be written $r^2=c^4+d^4$.

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