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?
 A: 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!
A: I tried the heuristic from the Hardy-Littlewood circle method for this equation. The heuristic suggests that the number of solutions within the range $\max\{\vert x\vert,\vert y\vert,\vert z\vert\}<N$ should be something like
$$\sigma_{\infty}\prod_{p}\sigma_p,$$
where $\sigma_{\infty}$ and $\sigma_{p}$ are real density and local densities.
$$\sigma_{\infty}=\lim_{\epsilon\rightarrow 0}\frac{1}{2\epsilon}\vert\{(x,y,z)\mid\max\{\vert x\vert,\vert y\vert,\vert z\vert\}<N,\vert x^4+y^4+1-z^2\vert<\epsilon\}\vert$$
$$\sigma_p=\lim_{n\rightarrow \infty}\vert\{(x,y,z)\mid x^4+y^4+1\equiv z^2\mod p^n\}\vert/p^{2n}$$
These densities have explicit formulas(if I calculated them correctly)
if $p\equiv 3 \mod 4$,
$$\sigma_p=1-1/p.$$
if $p\equiv 1 \mod 4$,
$$\sigma_p=1+\frac{1+6(-1)^{(p-1)/4}}{p}-\frac{2a}{p^2},$$
where $p=a^2+b^2$, $a\equiv 3\mod 4$.
$\sigma_2=1$ and $$\sigma_{\infty}\approx\frac{B(1/4,1/4)}{2}\log 2N\approx 3.70815\log 2N,$$ where $B$ is Euler beta function.
The infinite product over prime numbers $p$ is not absolutely convergent, but it is indeed convergent. 
$$\prod_{p}\sigma_p\approx 0.0193327$$.
$(\pm x,\pm y,\pm z)$ and $(\pm y,\pm x,\pm z)$ are all solutions of the Diophantine equation, so the number of essentially different solutions should be $$\frac{B(1/4,1/4)}{32}\prod_{p}\sigma_p\cdot\log 2N\approx 0.0044805\log 2N.$$ So, if this heuristic works, the first solution may occur near $N\sim\exp(1/0.0044805)/2\approx 4\times 10^{96}$, which means $x,y\sim 10^{48}$. It is possible that the solutions are enormous, but I am not sure if there are other obstructions to the equation.

Edit: I guess things are different for Euler's quartic $$x^4+y^4+z^4=w^4.$$
The real density for this surface is still about $c\log N$, but the infinite product over primes diverges to $\infty$. The infinite product grows like $(\log N)^r$, where $r$ is some real number. So I think the integral points on Euler's quartic is quite "dense" compared to $x^4+y^4+1=z^2$.
A: 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.
