Let $n,y$ be positive integers that are greater than 1. Suppose that $x^n=y$ does not have a solution in positive integers. Then is it true that for infinitely many primes $p$ there are no solutions $x^n\equiv y\pmod p$? It is known from here that if $f(x)$ is irreducible over $\mathbb{Q}$ then $f(x)\equiv 0\pmod p$ has no solution for infinitely many primes $p$. However, $f(x)=x^n-y$ is irreducible only under certain conditions. If the statement is true can you point towards a reference for a proof. I suspect that some higher reciprocity laws must be used here.
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7$\begingroup$ is it Maddest of them all or Maddest of the mall? $\endgroup$Will Jagy– Will Jagy2025-12-09 21:45:47 +00:00Commented Dec 9 at 21:45
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This is not true, and a famous counter-example is provided by the Grunwald-Wang theorem.
The counter-example is very simple. Take $n = 8$ and $y = 16$. Then $16$ is not an $8$th power but is an $8$th power modulo all odd primes. This follows from the relation $$1 = \left(\frac{4}{p}\right) = \left(\frac{2}{p}\right) \left(\frac{-2}{p}\right) \left(\frac{-1}{p}\right)$$ for all odd primes $p$, from which it follows that at least one of $-1,2,-2$ is a quadratic residue modulo all odd primes $p$, from where it is easy to deduce that $16$ is an $8$th power modulo all odd primes $p$.
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$\begingroup$ Thank you for your answer. Is there a "refined" statement for which the condition works? $\endgroup$Maddestofthemall– Maddestofthemall2025-12-10 08:55:08 +00:00Commented Dec 10 at 8:55
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3$\begingroup$ The Grunwald-Wang theorem gives a complete solution to this problem over general number fields. This is explained in the wikipedia article. You can also read more about it in the Chapter on the "Hasse principle" in the book "Neukirch, Schmidt, Wingberg - Cohomology of Number Fields". $\endgroup$Daniel Loughran– Daniel Loughran2025-12-10 10:46:13 +00:00Commented Dec 10 at 10:46