proof of Lucas-Lehmer primality test

The objective of this article is to prove the Lucas-Lehmer primality test:
Let $p>2$ be a prime, and let $M_p=2^p-1$ be the corresponding Mersenne number. Then $M_p$ is prime if and only if $M_p$ divides $s_{p-1}$ (equivalently, if and only if $s_{p-1}\equiv 0(M_p)$) where the numbers ${(s_n)}_{n\ge 1}$ are given by the following recurrence relation:

$s_1$$=4$

$s_{n+1}$$=s_n{}^2-2,n\ge 1$

We show that the validity of the primality test is equivalent to the following theorem, which is then proved directly:

Theorem 1.

(Lucas) $M_p$ is prime if and only if ${\alpha }^{\mathrm{(}{M}_p\mathrm{+}\mathrm{1}\mathrm{)}\mathrm{/}\mathrm{2}}\mathrm{\equiv }\mathrm{-}\mathrm{1}\mathrm{(}{M}_p\mathrm{)}$, where $\alpha \mathrm{=}\mathrm{2}\mathrm{+}\sqrt{\mathrm{3}}$.

To see that the two are in fact equivalent, let $\beta =2-\sqrt{3}$. Then $\alpha +\beta =4,\alpha \beta =1$. Thus

$s_1$$=\alpha +\beta$

$s_2$$={(\alpha +\beta )}^2-2={\alpha }^2+{\beta }^2+2\alpha \beta -2={\alpha }^2+{\beta }^2$

$s_3$$={\alpha }^4+{\beta }^4$

$\mathrm{\cdots }$

$s_{p-1}$$={\alpha }^{2^{p-2}}+{\beta }^{2^{p-2}}$

Note that $2^{p-2}=\frac{M_p+1}{4}$. Then

$s_{p-1}\equiv 0(M_p)$$\iff {\alpha }^{(M_p+1)/4}+{\beta }^{(M_p+1)/4}\equiv 0(M_p)$

$\iff {\alpha }^{(M_p+1)/2}+{(\alpha \beta )}^{(M_p+1)/4}\equiv 0(M_p)$

$\iff {\alpha }^{(M_p+1)/2}\equiv -1(M_p)$

It thus remains to prove Theorem 1. We start with two simple lemmas:

Lemma 2.

If $p\mathrm{>}\mathrm{3}$ is prime, then ${\alpha }^{p\mathrm{-}\mathrm{1}}\mathrm{\equiv }\mathrm{1}\mathrm{(}p\mathrm{)}$ or ${\alpha }^{p\mathrm{+}\mathrm{1}}\mathrm{\equiv }\mathrm{1}\mathrm{(}p\mathrm{)}$.

Proof.

where $\left(\frac{\cdot }{\cdot }\right)$ is the Legendre symbol. Thus

$$\left(\frac{3}{p}\right)=1\Rightarrow {\alpha }^{p-1}={\alpha }^p{\alpha }^{-1}={\alpha }^p\beta \equiv \alpha \beta =1(p)$$

$$\left(\frac{3}{p}\right)=-1\Rightarrow {\alpha }^{p+1}={\alpha }^p\alpha \equiv \beta \alpha =1(p)$$

∎

Lemma 3.

Let $p$ be a prime with $p\mathrm{\equiv }\mathrm{7}\mathrm{(}\mathrm{8}\mathrm{)}$ and $p\mathrm{\equiv }\mathrm{7}\mathrm{(}\mathrm{12}\mathrm{)}$. Then ${\alpha }^{\mathrm{(}p\mathrm{+}\mathrm{1}\mathrm{)}\mathrm{/}\mathrm{2}}\mathrm{\equiv }\mathrm{-}\mathrm{1}\mathrm{(}p\mathrm{)}$.

Proof.

${(1+\sqrt{3})}^2=4+2\sqrt{3}=2\alpha$, so that

$${(1+\sqrt{3})}^{p+1}=2^{(p+1)/2}{\alpha }^{(p+1)/2}$$

But $p\equiv 7(8)$, so that $\left(\frac{2}{p}\right)=1$. Thus $2^{(p+1)/2}\equiv 2\cdot 2^{(p-1)/2}\equiv 2(p)$ and therefore

$${(1+\sqrt{3})}^{p+1}\equiv 2{\alpha }^{(p+1)/2}(p)$$

Also,

$${(1+\sqrt{3})}^{p+1}=(1+\sqrt{3}){(1+\sqrt{3})}^p\equiv (1+\sqrt{3})(1+3^{(p-1)/2}\sqrt{3})(p)$$

But $p\equiv 7(12)$, so $3^{(p-1)/2}\equiv -1(p)$ and thus

$${(1+\sqrt{3})}^{p+1}\equiv (1+\sqrt{3})(1-\sqrt{3})=-2(p)$$

Putting together the two expressions for ${(1+\sqrt{3})}^{p+1}$, we get ${\alpha }^{(p+1)/2}\equiv -1(p)$.∎

We are now in a position to prove Theorem 1:

Proof.

$(\Rightarrow ):$ If $M_p$ is prime where $p>3$ is prime, then note that $M_p\equiv 7(8),7(1)2$ so that $M_p$ satisfies the conditions of Lemma 3. The result follows.

$(\Leftarrow ):$ If ${\alpha }^{(M_p+1)/2}\equiv -1(M_p)$, choose $q\mid M_p$ for $q$ a prime. Since $M_p\equiv 7(1)2$, we have $q>3$. Since ${\alpha }^{(M_p+1)/2}\equiv -1(M_p)$ also ${\alpha }^{(M_p+1)/2}\equiv -1(q)$ and thus ${\alpha }^{M_p+1}\equiv 1(q)$. But $M_p+1=2^p$, so

$${\alpha }^{2^p}\equiv 1(q)$$

Thus the order of $\alpha (q)$ divides $2^n$. It can’t divide $2^{n-1}$ since ${\alpha }^{(M_p+1)/2}\equiv -1(q)$, so its order is precisely $2^n=M_p+1$. However, ${\alpha }^{q+1}\equiv 1(q)$ or ${\alpha }^{q-1}\equiv 1(q)$ by Lemma 2 and thus $q\ge M_p$. But $q\mid M_p$, so $q=M_p$ and $M_p$ is in fact prime.∎