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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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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jun 4 '12 at 1:04
  • $\begingroup$ you are right. I edit it $\endgroup$ – Brhn Jun 4 '12 at 1:15
  • $\begingroup$ @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. $\endgroup$ – user14082 Aug 4 '12 at 16:34
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    $\begingroup$ See also: en.wikipedia.org/wiki/Fermat%27s_right_triangle_theorem $\endgroup$ – user940 Jul 21 '17 at 16:36
  • $\begingroup$ Related: math.stackexchange.com/questions/26548/… $\endgroup$ – Watson Nov 25 '18 at 15:24
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Suppose $z^2=y^4-x^4$ with $xyz\not=0$ for the smallest possible value of $y^4$. 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 $\left({z/p^2}\right)^2=\left({y/p}\right)^4-\left({x/p}\right)^4$ 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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  • $\begingroup$ Why? Other solutions could give extra insight. $\endgroup$ – user940 Jun 4 '12 at 1:22
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    $\begingroup$ Nope it was the same as yours. $\endgroup$ – user17762 Jun 4 '12 at 1:22
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    $\begingroup$ 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. $\endgroup$ – user236182 Nov 3 '15 at 10:59
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    $\begingroup$ @ByronSchmuland In the last sentence, why is that a smaller solution, you have to prove that $xy<z$, how do you prove that? The same question I will ask for the equation $v^4-u^4=w^2$, you didn't prove that $w<z$. $\endgroup$ – 76david76 Jun 30 '17 at 10:08
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    $\begingroup$ @76david76 You are quite right, there is a gap in the proof. I think that I have fixed it by changing the criterion for ordering solutions. Now I use $y$ instead of $z^2$. Please look it over and see if you agree that the solution is OK now. $\endgroup$ – user940 Jul 15 '17 at 13:05

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