Does the equation $x^4+y^4+1 = z^2$ have a non-trivial solution?

The background of this question is this: Fermat proved that the equation, $$x^4+y^4 = z^2$$

has no solution in the positive integers. If we consider the near-miss, $$x^4+y^4-1 = z^2$$

then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss, $$x^4+y^4+1 = z^2$$

has none with $0 < x,y < 10^6$.

Does the third equation really have none at all, or are the solutions just enormous?

Solution 1:

A related fun fact. Assuming a conjecture of Tyszka, then one "only" has to search up to $10^{38.532}$ to determine if the equation has finitely many solutions.

(Edit: Tyszka's conjecture is false, see http://arxiv.org/abs/1309.2682 for details.)

Express the equation as a restricted form system:

$x_1=1,$
$x_3=x_2*x_2,$
$x_4=x_3*x_3$ (here $x_2$ is the $x$ in the original equation),
$x_6=x_5*x_5,$
$x_7=x_6*x_6$ (here $x_5$ is the $y$),
$x_9=x_8*x_8,$ (here $x_8$ is the $z$)
$x_{10}=x_4+x_7,$
$x_{10}=x_9+x_1$ (this pulls it all together).

This requires 10 variables, so is a subset of Tyszka's system $E_{10}$. By Tyszka's conjecture, if this system has finitely many solutions in the integers then every solution has every variable assigned a value with absolute value at most $2^{2^{n-1}}=2^{512}$. Hence $|x|^4,|y|^4,|z|^2 \le 2^{512}$ so $|x|,|y|\le 2^{128}\lt 10^{38.532}$.

Note that there is a big gap between $10^6$ and the bound given by Tyszka's conjecture. Even if the conjecture does hold had held, there may be infinitely many solutions, but the smallest one may have $x,y>10^{38.53}$.

The message here is just that searching the interval from $0$ to $10^6$ isn't enough.

Solution 2:

Edit: The question asks about a different equation than I thought when I read it first time. I will leave this anyway because some people reading the question may be interested.

On the website http://sites.google.com/site/tpiezas/009 there is the identity listed:

$$(17p^2-12pq-13q^2)^4 + (17p^2+12pq-13q^2)^4 = (289p^4+14p^2q^2-239q^4)^2 + (17p^2-q^2)^4$$

(17*p^2-12*p*q-13*q^2)^4 + (17*p^2+12*p*q-13*q^2)^4 = (289*p^4+14*p^2*q^2-239*q^4)^2 + (17*p^2-q^2)^4

That checks out.

One can solve the Pell equation $q^2 - 17 p^2 = \pm 1$, e.g. (p,q) = (0,1),(1,4),(8,33),(65,268),... which lead to infinitely many solutions of the Diophantine equation:

• $13^4 + 13^4 = 239^2 + 1$
• $239^4 + 143^4 = 60671^2 + 1$
• $16237^4 + 9901^4 = 281275631^2 + 1$

I don't know whether that is all solutions! I suspect so. Thanks to Henry, this is not all solutions!