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

Question

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?

Answer 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.

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. And the conjecture may also not hold... Finally, even if the conjecture does hold, there may be infinitely many solutions, but the smallest one may have $x,y>10^{38.53}$ and $z>10^{77.06}$.

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

Answer 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!