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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?

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I assume you're talking about this: 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
up vote 10 down vote

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:

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

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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 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!

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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

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