1

Let $n$ be a positive integer greater than or equal to $3$, and let $k$ be an arbitrary but fixed nonzero integer. Are there infinitely many integer solutions to the equation $x^n + y^n =z^n + k$? So, for example, are there infinitely many solutions to $x^3 + y^3 = z^3 + 1$, etc. Also, does it depend on the precise value of $k$ and $n$, so that some $k$'s and $n$'s have infinitely many solutions, and others don't?

user107952
  • 20,508
  • For $k=1$ , it might be helpful to look up the near-misses. – Peter Aug 09 '23 at 17:48
  • 1
    Not sure whether Faltings theorem can be applied here to establish finite many solutions for most of the pairs. – Peter Aug 09 '23 at 17:50
  • In relation with https://math.stackexchange.com/questions/4745532/has-equation-anbn-cn1-infinite-solutions-with-n2/4745571#4745571 – Piquito Aug 09 '23 at 17:58
  • 2
    Consider any $n \ge 5$ where $2n+1$ is a prime. Then, by Fermat's little theorem since for all $a$ where $2n + 1 \nmid a$ we have $a^{2n}\equiv 1\pmod{2n+1}$, then $x^n$, $y^n$ and $z^n$ are all congruent to $-1$, $0$ or $1$ modulo $2n+1$. Thus, there will be no solutions for any $k \equiv m\pmod{2n+1}$ where $4 \le m \le 2n-3$. For example, with $n=5$, there will not be any solutions for $4 \le k \le 7$. – John Omielan Aug 09 '23 at 18:04
  • @Peter: Faltings works for curves, but this equation seems to define a surface… – Aphelli Aug 09 '23 at 20:35

1 Answers1

1

We can trivially check that $(n,k)=(3,1)$ has infinitely many Solutions :
$(x,y,z)=(c,1,c)$

More-over , we can then trivially check that $(n,k)=(3,-1)$ has infinitely many Solutions :
$(x,y,z)=(c,-1,c)$

Semi-non-trivially (?!) , we know that $(n,k)=(2m+1,\pm 1)$ has infinitely many Solutions :
$(x,y,z)=(c,\pm 1,c)$

Non-trivially (checking the modulo 9) , we know that $(n,k)=(9,3)$ has no Solutions.

Similarly , we know that $(n,k)=(24,5)$ has no Solutions.

More-over (checking the modulo 24) : $(n,k)=(24,3|4|5|6|11|12|13|14|19|20|21|22)$ : All have no Solutions.

Hence , we have to check it Case-By-Case.

I suspect there will be no easy "over-all" characterization.

Prem
  • 9,669