Integer solutions of x^n + y^n = z^{n-1}

2012-10-08T09:13:49Z

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 $\varphi^{-1}(n)$.

Answer by Ilya Bogdanov for Integer solutions of x^n + y^n = z^{n-1}

2012-10-08T10:01:15Z

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.

Integer solutions of x^n + y^n = z^{n-1} - MathOverflow

Integer solutions of x^n + y^n = z^{n-1}

Answer by Ilya Bogdanov for Integer solutions of x^n + y^n = z^{n-1}