If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$

Question

Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$.

Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it.

We have $x^2 \equiv -3 \pmod{p}$ for some integer $x$. By Fermat's Little Theorem we have $$x^{p-1} \equiv x^2 \cdot x^{p-3} \equiv -3 \cdot x^{p-3} \equiv 1 \pmod{p}.$$ Thus, $x^{p-3} \equiv -3^{-1} \pmod{p}$.

I didn't see how to get a contradiction if $p \equiv 2 \pmod{3}$.

Answer 1

Lemma: for an odd prime $q,$ we always have $$ (-3|q) = (q |3). $$ Proof: if $q \equiv 1 \pmod 4,$ then $$ (-3|q) = (-1|q)(3|q) = (3|q) = (q|3). $$ If $q \equiv 3 \pmod 4,$ then $$ (-3|q) = (-1|q)(3|q) = -(3|q) = (q|3). $$

In the lemma, $3$ can be replaced by any prime $r \equiv 3 \pmod 4.$

Let $n = 2k+1.$ We have $1 + 3^n = 1 + 3 (3^k)^2 = 1 + 3 w^2.$ If $p \equiv 2 \pmod 3$ and $$ x^2 + 3 y^2 \equiv 0 \pmod p, $$ assume $y$ is not divisible by $p.$ Then $y$ has a multiplicative inverse $\pmod p.$ $$ x^2 \equiv -3 y^2 \pmod p, $$ $$ \frac{x^2}{y^2} \equiv -3 \pmod p, $$ $$ \left( \frac{x}{y} \right)^2 \equiv -3 \pmod p. $$ This contradicts $p \equiv 2 \pmod 3,$ so actually $y$ is divisible by $p.$ It follows that $x$ is also divisible by $p.$

Back to the original problem, we have $p \equiv 2 \pmod 3$ and $$ 1 + 3 w^2 \equiv 0 \pmod p.$$ However, this implies $1$ is divisible by $p,$ which is false

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The easy proposition is this: we are given a quadratic form $$ a x^2 + b x y + c y^2 $$ with discriminant $$ \Delta = b^2 - 4 a c,$$ finally an odd prime $q$ such that $$ (\Delta | q) = -1. $$ IF $$ a x^2 + b x y + c y^2 \equiv 0 \pmod q $$ THEN both $$ x,y \equiv 0 \pmod q, $$ and we get the extra $$ a x^2 + b x y + c y^2 \equiv 0 \pmod {q^2} $$

Answer 2

This may qualify as cheating, but I will use a few known facts from Wikipedia. First of all, indeed $3^{n+1} \equiv -3 \pmod{p}$, where $n+1$ is even, thus $$\left(\frac{-3}{p}\right)=1 \Rightarrow \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=1 \tag{1}$$ But $$\left(\frac{-1}{p}\right)=\left\{\begin{matrix} 1, & p \equiv 1 \pmod{4} \\ -1,& p \equiv 3 \pmod{4} \end{matrix}\right. \color{red}{\text{ and }} \left(\frac{3}{p}\right)=\left\{\begin{matrix} 1, & p \equiv 1 \text{ or } 11 \pmod{12} \\ -1, & p \equiv 5 \text{ or } 7 \pmod{12} \end{matrix}\right.$$ Due to $(1)$, we have $$\left\{\begin{matrix} \left(\frac{-1}{p}\right)=1\\ \left(\frac{3}{p}\right)=1 \end{matrix}\right. \color{red}{\text{ or }} \left\{\begin{matrix} \left(\frac{-1}{p}\right)=-1\\ \left(\frac{3}{p}\right)=-1 \end{matrix}\right. $$ Or, for the first case $$\left\{\begin{matrix} \left(\frac{-1}{p}\right)=1\\ \left(\frac{3}{p}\right)=1 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} p \equiv 1 \pmod{4}\\ p \equiv 1 \text{ or } 11 \pmod{12} \end{matrix}\right. \Rightarrow p \equiv 1 \pmod{12}$$ and for the latter $$\left\{\begin{matrix} \left(\frac{-1}{p}\right)=-1\\ \left(\frac{3}{p}\right)=-1 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} p \equiv 3 \pmod{4}\\ p \equiv 5 \text{ or } 7 \pmod{12} \end{matrix}\right. \Rightarrow p \equiv 7 \pmod{12}$$ As a result $$p \equiv 1 \pmod{12} \color{red}{\text{ or }} p \equiv 7 \pmod{12} \Rightarrow \\ p \equiv 1 \pmod{3} \color{red}{\text{ or }} p \equiv 7 \pmod{3} \Rightarrow \\ p \equiv 1 \pmod{3}$$

Answer 3

You have shown that $(\frac{-3}{p}) = 1$ in this case. The truth is, in fact, that $(\frac{-3}{p}) = 1 \iff p \equiv 1 \pmod{3}$, which is what you are trying to prove.

$(\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p})$ due to multiplicity of Legendre symbol.

From Euler's criterion, $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}$.

From QR, $(\frac{3}{p}) = (-1)^{\frac{p-1}{2}}(\frac{p}{3})$.

Hence $(\frac{-3}{p}) = (-1)^{p-1}(\frac{p}{3}) = (\frac{p}{3}) = 1 \iff p \equiv 1 \pmod{3}$.