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As a generalisation to the equation of Fermat, one can ask for rational solutions of $X^n+Y^n+Z^n=1$ (or almost equivalently integer solutions of $X^n+Y^n+Z^n=T^n$).

Contrary to the case of Fermat, the case where $n=3$ has infinitely many solutions, because the surface is rational. For $n=4$, we get a K3 surface and for $n\ge 5$ the surface should have finitely many points or the points should be contained in finitely many curves since it is of general type (Bombieri-Lang conjecture).

But are there some non-trivial solutions for $n\ge 5$ (and $n=4$)? Do we know if there are finitely many solutions for $n\ge 5$ ?

I guess that it should be classical, but I did not find it on this site or online, after googling "Fermat, surface, rational solutions", etc

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    $\begingroup$ In the integer case, with $n=4$ there is some info here, on a disproved conjecture of Euler, en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture Elkies had found examples. For $n\geq 5$ examples are known if one uses more variables. (I guess with 3 summands, $n\geq 5$ nothing is known) $\endgroup$
    – Christian Elsholtz
    Commented Dec 14, 2015 at 18:48
  • $\begingroup$ For $n=4$, there is some updated data on, "More elliptic curves for $x^4+y^4+z^4 = 1$" in this post. $\endgroup$
    – Tito Piezas III
    Commented Dec 15, 2015 at 3:44
  • $\begingroup$ Also, for $n=5$, Desboves gave a nice solution in terms of complex numbers as, $$(a\sqrt{2}+b)^5 + (-b+c\sqrt{-2})^5 = (a\sqrt{2}-b)^5 + (b+c\sqrt{-2})^5$$ where $a^2+b^2 = c^2$. $\endgroup$
    – Tito Piezas III
    Commented Dec 16, 2015 at 5:56

5 Answers 5

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It has been conjectured by Euler that this equation has no solutions in positive integers when $n\geq 4$.

When $n=4$, this was disproved by Elkies in the paper [Elkies, On A4+B4+C4=D4] in a very strong way: he proves that the rational points of this K3 surface are dense in the real points for the euclidean topology.

When $n\geq 5$ is odd, your surface contains lines, for instance the line $Z-T=X+Y=0$. Consequently, it has infinitely many rational points. Of course, this does not disprove Euler's conjecture, that required positive integers.

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Here are some references:

MR2077618 Gundersen, Gary G.; Tohge, Kazuya Entire and meromorphic solutions of $f^5+g^5+h^5=1.$ Symposium on Complex Differential and Functional Equations, 57–67, Univ. Joensuu Dept. Math. Rep. Ser., 6, Univ. Joensuu, Joensuu, 2004. https://www.researchgate.net/publication/260518318_Entire_and_meromorphic_solutions_of_f5_g5_h5_1

MR1821651 Gundersen, Gary G. Meromorphic solutions of $f^5+g^5+h^5≡1.$ Complex Variables Theory Appl. 43 (2001), no. 3-4, 293–298.

MR1660942 Gundersen, Gary G. Meromorphic solutions of $f^6+g^6+h^6≡1.$ Analysis (Munich) 18 (1998), no. 3, 285–290.

They study not only rational but also entire and meromorphic in the plane solutions, and they mention in these papers what is known on the subject. There is also a survey:

MR3170744 Hayman, W. K. Waring's theorem and the super Fermat problem for numbers and functions. Complex Var. Elliptic Equ. 59 (2014), no. 1, 85–90.

According to the very recent preprint of Gundersen, http://arxiv.org/abs/1509.02225

Equation $f^n+g^n+h^n=1$ has solutions in rational functions for $n\leq 5$ and no such solutions for $n\geq 8$. The cases $n=6, n=7$ are open. Non-constant rational solutions for $n=5$ are in the first article above.

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  • $\begingroup$ Umm, $f^n+g^n+h^n = 1$ has no known rational solution $fgh \neq 0$ when $n=5$. $\endgroup$
    – Tito Piezas III
    Commented Dec 15, 2015 at 3:38
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    $\begingroup$ I edited for clarity. In my answer, "rational" means a non-constant element of $C(z)$ :-) $\endgroup$
    – Alexandre Eremenko
    Commented Dec 15, 2015 at 3:52
  • $\begingroup$ Ok, I follow these things, and I thought I somehow missed that big news. :) $\endgroup$
    – Tito Piezas III
    Commented Dec 15, 2015 at 3:54
  • $\begingroup$ I don't follow these things, just cited what the preprint says. The exact reference is MR2077618. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 15, 2015 at 3:56
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For the equation.

$$x^5+y^5+z^5=w^5$$

You can write such a simple solution.

$$x=(p^2+s^2)\sqrt{-1}+p^2+2ps-s^2$$

$$y=(p^2+s^2)\sqrt{-1}+s^2-2ps-p^2$$

$$z=p^2-2ps-s^2-(p^2+s^2)\sqrt{-1}$$

$$w=(p^2+s^2)\sqrt{-1}+p^2-2ps-s^2$$

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    $\begingroup$ This can also be given a Desboves-type form as, $$(a + b + c \,i)^5 + (-a - b + c \,i)^5 + (a - b - c \,i)^5 = (a - b + c \,i)^5$$ where $a^2+b^2 = c^2$. $\endgroup$
    – Tito Piezas III
    Commented Dec 16, 2015 at 8:55
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In the year 1988 Noam Elkies found $2682440^4+15365639^4+18796760^4=20615673^4$.

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I found a parameterized solution of $X^5+Y^5+Z^5=T^5$ with $X$ and $T$ rational and $Y$ and $Z$ complex rational:

$(k^2-4k+1)^5+(2k-2+(k^2-2k+3)i)^5+(2k-2-(k^2-2k+3)i)^5=(k^2-3)^5$

There is an almost Desboves form of it, as follows:

$(b-a)^5+(a+ci)^5+(a-ci)^5=(b+a)^5$, where $2a^2+b^2=c^2$

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