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I have a question

show that $x^4- y^4 = z^2$ has no nontrivial solution where $x$, $y$ and $z$ are nonzero integers

I tried infinite descent to find solution but I could not find it. square of a number in mod 4 is 1 or 0 I also tried to use but got nothing.

Can you help? thanks

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This equation has infinitely many solutions, for example, $x=y=17$, $z=0$. You'll have a better chance of solving the problem if you get the statement right. – Gerry Myerson Jun 4 '12 at 1:04
you are right. I edit it – Brhn Jun 4 '12 at 1:15
@Brhn Please accept the answer if the answer is correct and you are satisfied with it, else please let us know if you have some more doubts. – Jayesh Badwaik Aug 4 '12 at 16:34

Suppose $z^2=y^4-x^4$ with $xyz\not=0$ for the smallest possible value of $z^2$. First we rewrite the equation as $y^4=x^4+z^2$ so that $\{z,x^2,y^2\}$ is a Pythagorean triple. It must be primitive, since if some prime $p$ divides $\gcd(x^2,y^2)$, then $p\,|\,y^2$ implies $p\,|\,y$ which gives $p^4\,|\,y^4$. Similarly, $p^4\,|\,x^4$, so $p^4\,|\,z^2$. This implies $p^2\,|\,z$, so that $\left({y/p}\right)^4-\left({x/p}\right)^4=\left({z/p^2}\right)^2$ is a smaller solution.

The Pythagorean triple $z,x^2,y^2$ is primitive and there are two cases:

If $x$ is even, then for some $m>n$, $(m,n)=1$, and $m\not\equiv n \pmod2$ we have $$ z=m^2-n^2,\quad x^2=2mn,\quad y^2=m^2+n^2.$$ Since $m,n$ have opposite parity, we can let $o$ denote the odd number and $e$ the even number among $\{m,n\}$. The primitive Pythagorean triple $\{n,m,y\}$ gives $$o=t^2-s^2,\quad e=2st,\quad y=t^2+s^2,$$ for some $t>s$, $(t,s)=1$, and $t\not\equiv s\pmod2$. The formula for $x^2$ now gives $$(x/2)^2=ts(t^2-s^2)$$ which expresses the product of three relatively prime numbers as a square. That means all three of them are squares: $s=u^2$, $t=v^2$, and $t^2-s^2=w^2$. In other words, $v^4-u^4=w^2$ is another solution to our equation, and it is smaller, since $v^4<t^2+s^2=y\leq y^4$.

If $x$ is odd, then for some $m>n$, $(m,n)=1$, and $m\not\equiv n\pmod2$ we have $$ x^2=m^2-n^2,\quad z=2mn,\quad y^2=m^2+n^2.$$ In this case $m^4-n^4=(xy)^2$ is a smaller solution, since $m^4<(m^2+n^2)^2=y^4$.

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+1. I had to delete my solution :). – user17762 Jun 4 '12 at 1:19
Why? Other solutions could give extra insight. – Byron Schmuland Jun 4 '12 at 1:22
Nope it was the same as yours. – user17762 Jun 4 '12 at 1:22
Oh, OK. ${}{}{}{}$ – Byron Schmuland Jun 4 '12 at 1:23
In the first paragraph, the Pythagorean triple doesn't have to be primitive just because extracting common divisors gives a smaller solution. But we can wlog let it be primitive. – user236182 Nov 3 '15 at 10:59

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