# $x^4 - y^4 = 2z^2$ 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.

• Are $x,y,z$ integers? Mar 22, 2013 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. Mar 22, 2013 at 18:55
• apparently $x,y,z$ are integers.
– user63181
Mar 22, 2013 at 18:56
• I assume $x$ and $y$ must be distinct, or you have a solution. Mar 22, 2013 at 19:00
• yah x,y,z are integers Mar 22, 2013 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.

• Fermat would be proud of you. Mar 22, 2013 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?
– user940
Mar 23, 2013 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 $2Z^2$. Mar 23, 2013 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.
– user940
Mar 23, 2013 at 14:41

IDENTITY--The identity $$(X^4 - Y^4)^2 + 4(XY)^4 = (X^4 + Y^4)^2$$ solves the proposer's question.

The proposer assumes that there is a non-zero triple of integers X,Y,Z

such that $$(X^4 - Y^4 ) = 2(Z^2)$$. WOLOG we may assume $$(X,Y,Z) =1$$.

We substitute " $$2(Z^2)$$ " for $$(X^4 -Y^4)$$ in the above IDENTITY.

This translates to $$(2Z^2)^2 + 4(XY)^4 = (2B)^2$$ where $$X^4 + Y^4 = 2B$$. ( Note that X,Y, are both odd as $$X^4 = Y^4 = 2(Z^2)$$ by hypothesis. Hence we may write $$X^4 + Y^4) = 2B$$ for some nonzero integer $$B$$ in the above IDENTITY.)

Dividing both sides by $$4$$ yields

$$Z^4 + (XY)^4 = B^2$$, which Fermat proved is impossible in non-zero integers $$Z$$,$$(XY)$$, and $$B$$ and is the connection that the proposer was seeking.

Many thanks to JW Tanner and Tinlyx for improving the formatting of this proof. I wanted to share the above IDENTITY equation with the readership because it is an important tool to shorten and simplify Fermat-type equations of exponent $$4$$ which may or may not have a "$$2$$" as a coefficient. The first answer given to this problem above is correct, but this solution is also very elegant, and simpler as well.