This is related to another question

I am interested in the non-trivial integer solutions of $$ x^n + y^n = z^{n-1} $$

for $n \ge 4$. A solution is trivial if $xyz=0$ or $x = \pm y$. There are infinitely many rational solutions to $x^n + y^n = (x+y)^{n-1}$ parametrized in the linked question.

For $n=5$ parametric solutions are $(-121 \cdot 2^{{\left(4 k + 3\right)}}, 363 \cdot 2^{{\left(4 k + 3\right)}},11^3 \cdot 2^{5k+4}) $

For $n > 5$ couldn't find any solution so far.

Q1. Are there non-trivial solutions for $n > 5$?

Q2. Are there $n$ for which non-trivial solutions don't exist?

Q3. Is it possible for some $n > 5$ to find solutions without searching? Parametrizing all solutions (this might settle a case of Fermat-Catalan Conjecture)?

Computationally the fastest way I found so far is pari's "t=thueinit(x^5+1,1);sol=thue(t,a^4);" though iterating over the divisors is another option. There are congruence conditions mod $\varphi^{-1}(n)$.


1 Answer 1


Take any $a,b$ and set $c=a^n+b^n$. Then the triple $(ac^{n-2},bc^{n-2},c^{n-1})$ is a solution of your equation.

Conversely, if $(x,y,z)$ is a solution and $d$ is their gcd, so $(x,y,z)=(ad,bd,cd)$, then you get $d(a^n+b^n)=c^{n-1}$. One of the solutions is presented above (with $d=c^{n-2}$). But there also exist smaller solutions --- they appear as soon as $a^n+b^n$ is not square-free.

  • 3
    $\begingroup$ One way to figure out that something like this might work is to projectivize the equation; it becomes $X^n + Y^n = WZ^{n-1}$ and now it is quite noticeable that the equation is linear in the new variable. $\endgroup$ Commented Oct 11, 2012 at 6:48
  • $\begingroup$ Thank you. I suppose this approach won't work for exponents (n,n,n-2) ? $\endgroup$
    – joro
    Commented Oct 13, 2012 at 12:37
  • $\begingroup$ No, it does not lead to an answer immediately. But perhaps it may simplify something. At least, it shows that for even $n$ we need to solve the equation $a^n+b^n=f^2$ (which seems to have no nontrivial solutions at least for $n=4$;) ). For odd $n$, although, it provides a solution $(ac^{(n-3)/2},bc^{(n-3)/2},c^{(n-1)/2})$, where again $c=a^n+b^n$. $\endgroup$ Commented Oct 13, 2012 at 21:19
  • $\begingroup$ Interesting :-). A result on Fermat Catalan conjecture states that (n,n,2) has no coprime solutions. $\endgroup$
    – joro
    Commented Oct 15, 2012 at 15:06
  • $\begingroup$ Would you please explain about the smaller solutions when not square-free? There are no nontrivial solutions known to $a^5+b^5=c^5+d^5$ and only two primitive known to $a^4+b^4=k$. $\endgroup$
    – joro
    Commented May 8, 2014 at 13:08

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