Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Duplicate math.stackexchange.com/questions/26591/… –  P.. Mar 24 '13 at 7:15
4  
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
add comment

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.