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Is it possible to show that for given $m$ and $k$, number of primes $p$ for which exists $n$ $( satisfying: $n^m + k\equiv 0\pmod{p}$ $(n+1)^m + k\equiv 0\pmod{p}$ is bounded (finite)?

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    @Gerry: Ah, you're right.2011-08-08

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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.

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    @Ante, if you want, you could always post a new question about the theory. But maybe you'd rather try to work it out yourself, from the literature on resultants.2019-04-22