# 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
From the faq: Please don't use signatures or taglines in your posts. Every post you make is already "signed" with your standard user card, which links directly back to your user page. Your user page belongs to you — fill it with interesting information about your interests, links to cool stuff you've worked on, or whatever else you like! –  Rasmus Jan 9 '11 at 15:43
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.

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... (Edit: the conjecture is false, see http://arxiv.org/abs/1309.2682 for details.) Finally, even if the conjecture does hold had held, there may be infinitely many solutions, but the smallest one may have $x,y>10^{38.53}$ and $z>10^{77.06}$.

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

-
What a coincidence! I just posted a question that this is answering! –  quanta Apr 16 '11 at 14:14

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