# Can $\ a^k+b^k+c^k\$ be prime for $\ k\in\{0,\ldots, m\}$ arbitary large?

Let $$\ a,b,c\$$ be integers with $$\ 0 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.

• The hypotehsis that prime-number question always receive a downvote is still alive ! – Peter Feb 17 at 11:32
• 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 $\sup_{(a,b,c)}f(a,b,c)=\infty$. – Samrat Mukhopadhyay Feb 17 at 11:41
• @SamratMukhopadhyay This is not what I mean. I mean whether we can find a triple $\ (a,b,c)\$ for every given $\ m\$ – Peter Feb 17 at 11:43
• 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 $0\le k\le m-1$? – Samrat Mukhopadhyay Feb 17 at 11:49
• Is it clear that the corresponding "two-variate" function (defined for $a^k+b^k$) is bounded? – W-t-P Feb 17 at 20:47