# $x^4 - y^4 = 2z^2$ has no solution

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.

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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

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.

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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