Intereting Posts

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prove or supply counter example about graph
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How to solve this system of linear equations
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What are the three non-isomorphic 2-dimensional algebras over $\mathbb{R}$?

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,

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

- Proposition 5.21 in Atiyah-MacDonald
- Order of Cyclic Subgroups
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- Any ring is integral over the subring of invariants under a finite group action
- Faithful irreducible representations of cyclic and dihedral groups over finite fields
- Transcendental Extensions. $F(\alpha)$ isomorphic to $F(x)$

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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- Orders of Elements in Minimal Generating sets of Abelian p-Groups
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- Annihilator of quotient module M/IM
- Is the group isomorphism $\exp(\alpha x)$ from the group $(\mathbb{R},+)$ to $(\mathbb{R}_{>0},\times)$ unique?
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- Image of a normal subgroup under a homomorphism
- System of Diophantine Equations

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.

**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!

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

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