# Number of solutions for pair of discrete logarithm like equations?

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

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 Neat question! I'm sure you've already worked out these cases, but just to mention them: For $m=1$, there are obviously no solutions for any $k$. For $m=2$, $n^2\equiv(n+1)^2\bmod p$ implies $2n+1\equiv 0\bmod p$, to which there is only one solution $n ## 1 Answer The two congruences will have a solution if the resultant (q.v.) of the polynomials$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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 Still I'm confused about theory background, but it works :-) It is easy to find possible solutions, just check prime factors of Sylvester matrix determinant. – Ante Aug 8 '11 at 19:38 @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. – Gerry Myerson Aug 9 '11 at 3:39