# $x^4+y^4=2z^2$ has only solution, $x=y=z=1$ .

How do I verify that the only solution in relatively prime positive integers of the equation $x^4+y^4=2z^2$ is $x=y=z=1$?

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–  P.. Mar 24 '13 at 7:15
It's not quite duplicate. That one asks how to solve it - and someone asserts that x=y=z=1 is the unique coprime solution. This question asks how to verify that it's the unique coprime solution. –  Glen O Mar 24 '13 at 7:31

Lemma I: If $(a,b,c)$ satisfies $a^2+b^2=c^2$, then $ab/2$ is not a square, nor twice a square.

Proof: By Euclid's formula, we can write $a=2pq$, $b=p^2-q^2$ $c=p^2+q^2$, so that $ab/2=pq(p+q)(p-q)$. If this is a square, then, as the four factors are pairwise coprime, there are $x,y,u,v$ such that $p=x^2$, $q=y^2$, $p+q=u^2$, and $p-q=v^2$. Then $2y^2=u^2-v^2=(u+v)(u-v)$. Since the g.c.d. of $u$ and $v$ is $2$, we conclude that one of $u+v,u-v$ is of the form $2r^2$, the other of the form $4s^2$. Hence $x^2=\frac{u^2+v^2}{2}=r^4+4s^4$ is another pythagorean triple with hypotenuse less than the original one, as $x^2=p^2<p^2+q^2$. So this completes the proof by descent.

On the other hand, if $ab/2$ is twice a square, then either $p=2x^2, q=y^2$, or $p=x^2, q=2y^2$. Moreover, $p+q=u^2$, and $p-q=v^2$, as above, with $u,v$ odd. Now $2p=\frac{u^2+v^2}{2}$, so $p$ must be odd. So $p=x^2$ and $q=2y^2$. Then $4y^2=2q=(u+v)(u-v)$. Thus $u+v=2r^2$ and $u-v=2s^2$ for some $r$ and $s$. Consequently $u=r^2+s^2$ and $v=r^2-s^2$. Finally, $x^2=p=\frac{u^2+v^2}{2}=r^4+s^4$ is another triple with area $2(\frac{rs}{2})^2$ and strictly less hypotenuse, hence again completing the proof.

Lemma II: No right-angled triangle with sides $a, b$, and hypotenuse $c$ can have $a,b$ both squares, nor $a,c$ both squares.

Proof: Suppose both $a$ and $b$ are squares, then we know that one of $a$ is divisible by $4$, and another odd, so that the area $ab/2$ is twice a square, contradicting the above lemma.
Suppose that $a$ and $c$ both are squares, then $a, c$ satisfy the equation $x^4-y^4=z^2$. If this equation has a non-trivial solution($x\neq y$), then, taking $p=x^2$ and $q=y^2$, we form a pythagorean triple $(2pq,p^2-q^2,p^2+q^2)$ with area $(xyz)^2$, again a contradiction.

Now your result follows easily: consider the triangle $(x^2z^2,\frac{x^4-y^4}{2},\frac{x^4+y^4}{2})$. By lemma II, neither $x^4+y^4$, nor $x^4-y^4$ can be twice a square, unless one is $0$, that is, $x=y=z=1$, which does not form a triangle. Q.E.D.

Ambiguity is never intended; the length is indeed a pain; so feel free to suggest improvements. Thanks in advance.

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Let me comment that this proof might be traced back to Euler the master! –  awllower Mar 24 '13 at 17:28