Is it possible to show that for given $m$ and $k$, number of primes $p$ for which exists $n$ $(
Is it possible to show that for given $m$ and $k$, number of primes $p$ for which exists $n$ $(
The two congruences will have a solution if the resultant (q.v.) of the polynomials $x^m+k$ and $(x+1)^m+k$ is a multiple of $p$. For fixed $m$, that resultant is a polynomial in $k$ of degree no more than $m$, so for fixed $k$ it only has a finite number of prime divisors. Thus there will be only a finite number of primes for which the congruences will have a solution.
$p\neq 2$
), namely $n=\frac{p-1}{2}$. Thus, there are no solutions unless $k=\frac{1}{4}\bmod p$, in which case the sole solution is $n=\frac{p-1}{2}$. The general case probably won't succumb to simple analysis like this though :) – 2011-08-07