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?

• I assume you're talking about this: members.bex.net/jtcullen515/Math10.htm? I would guess that the reason for the computer search is that it's an open problem that nobody knows the answer to. – Hans Lundmark Jan 9 '11 at 15:33
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• Yes, Hans, that is his site. Eq.2 can be solved via a Pell equation, so it has an infinity of positive integer solutions. Cullen and I tried to find parametrizations for p^4+q^2+1 = r^2, in the hope that we could specialize "q", but the easy identity I found had a "q" that was never a square. – Tito Piezas III Jan 9 '11 at 16:58
• @user9176: this is easily extended to conclude $10|x,y$ and that $z \equiv \pm 1 \text { or } \pm 1249 \mod 5000$. – Henry Apr 15 '11 at 11:40
• – Joel Reyes Noche Aug 28 '13 at 4:57

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.

• Can Tyszka's be true? Why does it not immediately contradict the solution to Hilbert's 10th problem, since one could just verify any Diophantine equation with no solutions has none by checking up to $2^2^{n-1}$ in this way? – JoshuaZ Sep 19 at 16:05

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!

• There is a list of smallish solutions to $x^4+y^4=z^2+1$ at the site linked by Hans Lundmark. The smallest is $5^4+7^4=55^2+1$ – Henry Apr 15 '11 at 11:31

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 $x^4+y^4+1=z^2$.

• +1 Can you make a small edit and apply your analysis to $a^4+b^4+c^4=d^4$ so we can compare its predicted lower bound with the actual smallest solution found by R. Frye? – Tito Piezas III Jul 22 '16 at 16:21
• @TitoPiezasIII: I think $a^4+b^4+c^4=d^4$ is different from $x^4+y^4+1=z^2$. I tried the same heuristic for this surface, and the infinite product diverges to $\infty$! So I guess there are more integral points on the surface $a^4+b^4+c^4=d^4$. – zy_ Jul 23 '16 at 13:31
• Maybe the question should be formulated differently? Find a parameterization of the solutions of this equation. $x^4+y^4=z^2-q^2$ And then find when there are solutions? $q=\pm1$ – individ Jul 24 '16 at 11:55