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6 votes
2 answers
425 views

About the solutions of $ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $

I already posted this question on MSE. Using theorem $IV$ from this article, it is possible to prove that when $p$ is a prime such that $p ≡ 3\bmod4$, $x ≢ y\bmod{p}$ and $\gcd(x,y) = 1$, then the ...
11 votes
2 answers
615 views

Jacobi symbols for two-square sums of primes

Given a prime $p\equiv 1\pmod 4$, Fermat's two-squares theorem discovered by Girard states that there exists two integers $A,B$ such that $p=A^2+B^2$. For all primes up to $10^7$ the integers $A$ and $...
7 votes
1 answer
339 views

Rational prime factors in the components of powers of Gaussian primes

Let $\pi=a+bi\in \mathbb{Z}[i]$ be a Gaussian prime with $a$ and $b$ nonzero, and $b$ even. For odd rational primes $p=\pi\bar\pi$ and $q\neq p$, define $\pi^{\frac{1}{2}\left(q-\left(\frac{-1}{q}\...
12 votes
1 answer
856 views

Quadratic reciprocity for three primes?

The quadratic reciprocity law states that for $p_1\ne p_2$ prime, the product $\left(\frac{p_1}{p_2}\right)\left(\frac{p_2}{p_1}\right)$ takes values $1$ or $-1$ depending on whether $p_1$ and $p_2$ ...
1 vote
0 answers
477 views

Legendre Symbol of a Very, Very Large Value

I'm trying to use FLINT (Fast Library for Number Theory) to calculate the Legendre Symbol of the following: $$\left(\frac{n! + 1}{p}\right)$$ In my case, $p$ is a positive, odd prime (specifically $...