Answer to Question #272614 in Combinatorics | Number Theory for Prathibha Rose

Question

Determine those odd primes p for which -3p=1 and those for which -

3p=-1

Answer 1

$\begin{aligned} &\text { Fermat's Little theorem states that if } \operatorname{gcd}(a, n)=1 \text { then } a^{\psi(n)} \equiv 1 \quad(\bmod \mathrm{n}) \\ &\text { The way we will use here to prove prove if } a^{2(p-1)} \neq 1 \quad(\bmod 3 \mathrm{p}) \text { then } \operatorname{gcd}(3 p, a)>1 \text { is prove by } \\ &\text { contradiction. } \\ &\text { Let } a^{2(p-1)} \not \equiv 1 \quad(\bmod 3 \mathrm{p}) \text { and } \operatorname{gcd}(3 p, a)=1 \\ &\text { As } \operatorname{gcd}(3 p, a)=1 \text { then by Fermat's Little theorem } \\ &a^{\psi(3 p)} \equiv 1 \quad(\bmod 3 \mathrm{p}) \\ &\Rightarrow a^{i(3) \psi(p)} \equiv 1 \quad(\bmod 3 \mathrm{p}) \\ &\Rightarrow a^{2(p-1)} \equiv 1 \quad(\bmod 3 \mathrm{p}) \\ &\text { a contradiction to } a^{2(p-1)} \not \equiv 1 \quad(\bmod 3 \mathrm{p}) \\ &\text { Hence } g \mathrm{~cd}(3 p, a)>1 \end{aligned}$