Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I prove that the equation $x^4 - y^4 = 2 z^2$ has no solutions using the fact that the equations $x^4 + y^4 = z^2$ and $x^4 - y^4 = z^2$ have no solutions. I cant think of a method of reducing the above equation to one of these forms.

share|cite|improve this question
Are $x,y,z$ integers? – Jeremy Mar 22 '13 at 18:55
Can you substitute $z = \frac{z'}{\sqrt{2}}$ for $z$ in the first equation? Then you have an equation of the form $x^4+y^4=z^2$. Apologies if this is in error - I can't remember if this is valid. – SSumner Mar 22 '13 at 18:55
apparently $x,y,z$ are integers. – user63181 Mar 22 '13 at 18:56
I assume $x$ and $y$ must be distinct, or you have a solution. – ncmathsadist Mar 22 '13 at 19:00
yah x,y,z are integers – noddy Mar 22 '13 at 19:01
up vote 10 down vote accepted

I think you meant nonexistence of positive solutions. Suppose that there exist some positive solutions(meaning that all of $x$, $y$, $z$ are positive). Then there is a positive solution $(x_0,y_0,z_0)$ with $x_0$ smallest.

First, notice that parity of $x_0$ and $y_0$ cannot be different. So, either both even or both odd. Both even case is not possible, because otherwise $(x_0/2,y_0/2,z_0/4)$ is a positive solution solution with smaller $x_0$.

Similarly with odd prime $p$, suppose $p|gcd(x_0,y_0)$, then you also get smaller solution $(x_0/p, y_0/p, z_0/p^2)$. Thus we can assume $(x_0,y_0)=1$.

Thus we can now assume that $x_0$ and $y_0$ are both odd, and coprime. Then by looking at the expression $$ \frac{x_0^2-y_0^2}{2}\cdot \frac{x_0^2+y_0^2}{2}=\frac{z_0^2}{2} $$

The right side must be an integer, so it must be $2Z^2$ , and any prime $p$ cannot divide both $\frac{x_0^2-y_0^2}{2}$ and $\frac{x_0^2+y_0^2}{2}$(otherwise we would have some prime $p$ dividing both $x_0$ and $y_0$. Then we have $$ x_0^2-y_0^2 = u^2$$ , and $$x_0^2+y_0^2 = 2v^2$$ for some positive integers $u,v$.

Now we solve for the Pythagorian triple in the first equation. $$ x_0=s^2+t^2\\ y_0=s^2-t^2\\ u=2st $$ for some positive integers $s,t$.

Then $x_0^2+y_0^2=2(s^4+t^4)=2v^2$. Hence we obtain $s^4+t^4=v^2$. However, this cannot have positive integer solution.

share|cite|improve this answer
Fermat would be proud of you. – marty cohen Mar 22 '13 at 23:18
Why do we have $x_0^2-y_0^2 = u^2$? Couldn't we have $x_0^2-y_0^2 = 2u^2$ instead? – Byron Schmuland Mar 23 '13 at 3:47
We started from both $x_0$ and $y_0$ odd. So for all prime power divisor of $(x_0^2+y_0^2)/2$ is odd, and the powers are even in $2Z^2$. – i707107 Mar 23 '13 at 4:53
Oh I see. Using modulo 4, we see that $(x_0^2+y_0^2)/2$ is the odd factor, and $(x_0^2-y_0^2)/2$ is the even one. – Byron Schmuland Mar 23 '13 at 14:41

Your Answer


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.