Let $\ a,b,c\ $ be integers with $\ 0<a<b<c\ $ and let $\ m\ $ be the smallest non-negative integer such that $$a^m+b^m+c^m$$ is composite.

Define $$f(a,b,c)=m-1$$ Hence $\ f(a,b,c)\ $ is the largest non-negative integer $\ m\ $ such that $$a^k+b^k+c^k$$ is prime for $\ k=0,\cdots ,m\ $. Since for $\ k=0\ $, we get always $\ 3\ $ which is a prime, $\ f(a,b,c)\ $ is always non-negative.

Has $f(a,b,c)$ a maximum ?

I have found $$f(2, 186, 803)=8$$ with brute force, that means that $\ 2^k+186^k+803^k\ $ is prime for $\ k=0,1,2,3,4,5,6,7,8\ $ which is unbeaten , if $\ c\le 1000\ $ holds.

  • 1
    $\begingroup$ The hypotehsis that prime-number question always receive a downvote is still alive ! $\endgroup$ – Peter Feb 17 at 11:32
  • $\begingroup$ I have a gut feeling that a maximum should exist. Otherwise, there is a triple $(a,b,c)$ such that for all $k\in \mathbb{N}$, $a^k+b^k+c^k$ produces prime numbers! That is we have a machine which outputs only prime numbers. I maybe very wrong, but that seems too good to be true to me. However, it might be possible that $\sup_{(a,b,c)}f(a,b,c)=\infty$. $\endgroup$ – Samrat Mukhopadhyay Feb 17 at 11:41
  • 1
    $\begingroup$ @SamratMukhopadhyay This is not what I mean. I mean whether we can find a triple $\ (a,b,c)\ $ for every given $\ m\ $ $\endgroup$ – Peter Feb 17 at 11:43
  • $\begingroup$ I am confused, if $m$ is given then $f(a,b,c)$ is given, then the question Has $f(a,b,c)$ a maximum? does not make sense to me. Are you trying to ask whether there is a triple $(a,b,c)$ such that, for a given $m$, $a^{k}+b^k+c^k$ is prime for all integers $k$ such that $0\le k\le m-1$? $\endgroup$ – Samrat Mukhopadhyay Feb 17 at 11:49
  • 1
    $\begingroup$ Is it clear that the corresponding "two-variate" function (defined for $a^k+b^k$) is bounded? $\endgroup$ – W-t-P Feb 17 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.