Some notes about power residues modulo prime

preprint2022arXivOpen access

Abstract

Let qq be a prime. We classify the odd primes p≠qp\neq q such that the equation x2≡q(modp)x^2\equiv q\pmod{p} has a solution, concretely, we find a subgroup L4q\mathbb{L}_{4q} of the multiplicative group U4q\mathbb{U}_{4q} of integers relatively prime with 4q4q (modulo 4q4q) such that x2≡q(modp)x^2\equiv q\pmod{p} has a solution iff p≡c(mod4q)p\equiv c\pmod{4q} for some c∈L4qc\in\mathbb{L}_{4q}. Moreover, L4q\mathbb{L}_{4q} is the only subgroup of U4q\mathbb{U}_{4q} of half order containing −1-1. Considering the ring Z[2]\mathbb{Z}[\sqrt{2}], for any odd prime pp it is known that the equation x2≡2(modp)x^2\equiv 2\pmod{p} has a solution iff the equation x2−2y2=px^2-2y^2=p has a solution in the integers. We ask whether this can be extended in the context of Z[2n]\mathbb{Z}[\sqrt[n]{2}] with n≥2n\geq 2, namely: for any prime p≡1(modn)p\equiv 1\pmod{n}, is it true that xn≡2(modp)x^n\equiv 2\pmod{p} has a solution iff the equation Dn2(x0,…,xn−1)=pD^2_n(x_0,\ldots,x_{n-1})=p has a solution in the integers? Here Dn2(xˉ)D^2_n(\bar{x}) represents the norm of the field extension Q(2n)\mathbb{Q}(\sqrt[n]{2}) of Q\mathbb{Q}. We solve some weak versions of this problem, where equality with pp is replaced by 0(modp)0\pmod{p} (divisible by pp), and the "norm" Dnr(xˉ)D^r_n(\bar{x}) is considered for any r∈Zr\in\mathbb{Z} in the place of 22.

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