Lucas-Lehmer Test

From ProofWiki
Jump to: navigation, search

Theorem

Let $q$ be an odd prime.

Let $\sequence {L_n}_{n \mathop \in \N}$ be the recursive sequence in $\Z / \paren {2^q - 1} \Z$ defined by:

$L_0 = 4, L_{n + 1} = L_n^2 - 2 \pmod {2^q - 1}$

Then $2^q - 1$ is prime if and only if $L_{q - 2} = 0 \pmod {2^q - 1}$.


Proof

Consider the sequences:

$U_0 = 0, U_1 = 1, U_{n + 1} = 4 U_n - U_{n - 1}$
$V_0 = 2, V_1 = 4, V_{n + 1} = 4 V_n - V_{n - 1}$


The following equations can be proved by induction:

\((1):\quad\) \(\displaystyle V_n\) \(=\) \(\displaystyle U_{n + 1} - U_{n - 1}\)
\((2):\quad\) \(\displaystyle U_n\) \(=\) \(\displaystyle \frac {\paren {2 + \sqrt 3}^n - \paren {2 - \sqrt 3}^n} {\sqrt {12} }\)
\((3):\quad\) \(\displaystyle V_n\) \(=\) \(\displaystyle \paren {2 + \sqrt 3}^n + \paren {2 - \sqrt 3}^n\)
\((4):\quad\) \(\displaystyle U_{m + n}\) \(=\) \(\displaystyle U_m U_{n + 1} - U_{m - 1} U_n\)


Now, let $p$ be prime and $e \ge 1$.

Suppose $U_n \equiv 0 \pmod {p^e}$.

Then $U_n = b p^e$ for some $b$.

Let $U_{n + 1} = a$.

By the recurrence relation and $(4)$, we have:

\(\displaystyle U_{2 n}\) \(=\) \(\displaystyle b p^e \paren {2 a - 4 b p^e} \equiv 2 a U_n\) \(\displaystyle \pmod {p^{e + 1} }\)
\(\displaystyle U_{2 n + 1}\) \(=\) \(\displaystyle U_{n + 1}^2 - U_n^2 \equiv a^2\) \(\displaystyle \pmod {p^{e + 1} }\)

Similarly:

\(\displaystyle U_{3 n}\) \(=\) \(\displaystyle U_{2 n + 1} U_n - U_{2 n} U_{n - 1} \equiv 3 a^2 U_n\) \(\displaystyle \pmod {p^{e + 1} }\)
\(\displaystyle U_{3 n + 1}\) \(=\) \(\displaystyle U_{2 n + 1} U_{n + 1} - U_{2 n} U_n \equiv a^3\) \(\displaystyle \pmod {p^{e + 1} }\)


In general:

\(\displaystyle U_{k n}\) \(\equiv\) \(\displaystyle k a^{k - 1} U_n\) \(\displaystyle \pmod {p^{e + 1} }\)
\(\displaystyle U_{k n + 1}\) \(\equiv\) \(\displaystyle a^k\) \(\displaystyle \pmod {p^{e + 1} }\)

Taking $k = p$, we get:

\((5):\quad\) \(\displaystyle U_n \equiv 0 \pmod {p^e}\) \(\leadsto\) \(\displaystyle U_{n p} \equiv 0\) \(\displaystyle \pmod {p^{e + 1} }\)


Expanding $\paren {2 \pm \sqrt 3}^n$ by the Binomial Theorem, we find that $(2)$ and $(3)$ give us:

\(\displaystyle U_n\) \(=\) \(\displaystyle \sum_k \binom n {2 k + 1} 2^{n - 2 k - 1} 3^k\)
\(\displaystyle V_n\) \(=\) \(\displaystyle \sum_k \binom n {2 k} 2^{n - 2 k + 1} 3^k\)


Let us set $n = p$ where $p$ is an odd prime.

From Binomial Coefficient of Prime, $\dbinom p k$ is a multiple of $p$ except when $k = 0$ or $k = p$.

We find that:

\(\displaystyle U_p\) \(\equiv\) \(\displaystyle 3^{\frac {p - 1} 2}\) \(\displaystyle \pmod p\)
\(\displaystyle V_p\) \(\equiv\) \(\displaystyle 4\) \(\displaystyle \pmod p\)

If $p \ne 3$, then from Fermat's Little Theorem:

$3^{p - 1} \equiv 1 \pmod p$

Hence:

$\paren {3^{\frac {p - 1} 2} - 1} \times \paren {3^{\frac {p - 1} 2} + 1} \equiv 0 \pmod p$
$3^{\frac {p - 1} 2} \equiv \pm 1 \pmod p$


When $U_p \equiv -1 \pmod p$, we have:

$U_{p + 1} = 4 U_p - U_{p - 1} = 4 U_p + V_p - U_{p + 1} \equiv -U_{p + 1} \pmod p$

Hence $U_{p + 1} \equiv 0 \pmod p$


When $U_p \equiv +1 \pmod p$, we have:

$U_{p - 1} = 4 U_p - U_{p + 1} = 4 U_p - V_p - U_{p-1} \equiv -U_{p - 1} \pmod p$

Hence $U_{p - 1} \equiv 0 \pmod p$


Thus we have shown that:

$(6) \quad \forall p \in \mathbb P: \exists \map \epsilon p: U_{p + \map \epsilon p} \equiv 0 \pmod p$

where $\map \epsilon p$ is an integer such that $\size {\map \epsilon p} \le 1$.


Now, let $N \in \N$.

Let $m \in \N$ such that $\map m N$ is the smallest positive integer such that:

$U_{\map m N} \equiv 0 \pmod N$

Let $a \equiv U_{m + 1} \pmod N$.

Then $a \perp N$ because:

$\gcd \set {U_n, U_{n + 1} } = 1$

Hence the sequence:

$U_m, U_{m + 1}, U_{m + 2}, \ldots$

is congruent modulo $N$ to $a U_0, a U_1, a U_2, \ldots$.

Then we have:

$(7) \quad U_n \equiv 0 \pmod N \iff n = k \map m N$

for some integral $k$.

(This number $\map m N$ is called the rank of apparition of $N$ in the sequence.)


Now, we have defined the sequence $\sequence {L_n}$ as:

$L_0 = 4, L_{n + 1} = \paren {L_n^2 - 2} \pmod {\paren {2^q - 1} }$

By induction it follows that:

$L_n \equiv V_{2^n} \pmod {\paren {2^q - 1} }$


We have the identity:

$2 U_{n + 1} = 4 U_n + V_n$

So any common factor of $U_n$ and $V_n$ must divide $U_n$ and $2 U_{n + 1}$.

As $U_n \perp U_{n + 1}$, this implies that $\gcd \set {U_n, V_n} \le 2$.

So $U_n$ and $V_n$ have no odd factor in common.

So, if $L_{q - 2} = 0$:

\(\displaystyle U_{2^{q - 1} } = U_{2^{q - 2} } V_{2^{q - 2} }\) \(\equiv\) \(\displaystyle 0\) \(\displaystyle \pmod {\paren {2^q - 1} }\)
\(\displaystyle U_{2^{q - 2} }\) \(\not \equiv\) \(\displaystyle 0\) \(\displaystyle \pmod {\paren {2^q - 1} }\)


Now, if $m = \map m {2^q - 1}$ is the rank of apparition of $2^q - 1$, it must be a divisor of $2^{q - 1}$ but not of $2^{q - 2}$.

So $m = 2^{q - 1}$.


Now we prove that $n = 2^q - 1$ must therefore be prime.

Let the prime decomposition of $n$ be $p_1^{e_1} \ldots p_r^{e_r}$.

All primes $p_j$ are greater than $3$ because $n$ is odd and congruent to $\paren {-1}^q - 1 = -2 \pmod 3$.

From $(5), (6), (7)$ we know that $U_t \equiv 0 \pmod {2^q - 1}$, where:

$t = \lcm \set {p_1^{e_1 - 1} \paren {p_1 + \epsilon_1}, \ldots, p_r^{e_r - 1} \paren {p_r + \epsilon_r} }$

where each $\epsilon_j = \pm 1$.

It follows that $t$ is a multiple of $m = 2^{q - 1}$.

Let $\displaystyle n_0 = \prod_{j \mathop = 1}^r p_j^{e_j - 1} \paren {p_j + \epsilon_j}$.

We have:

$\displaystyle n_0 \le \prod_{j \mathop = 1}^r p_j^{e_j - 1} \paren {p_j + \frac {p_j} 5} = \paren {\frac 6 5}^r n$

Also, because $p_j + \epsilon_j$ is even, $t \le \frac {n_0} {2^{r - 1} }$, because a factor of $2$ is lost every time the LCM of two even numbers is taken.

Combining these results, we have:

$m \le t \le 2 \paren {\frac 3 5}^r n \le 4 \paren {\frac 3 5}^r m < 3 m$

Hence $r \le 2$ and $t = m$ or $t = 2 m$, a power of $2$.

Therefore $e_1 = 1$ and $e_r = 1$.

If $n$ is not prime, we must have:

$n = 2^q - 1 = \paren {2^k + 1} \paren {2^l - 1}$

where $\paren {2^k + 1}$ and $\paren {2^l - 1}$ are prime.

When $q$ is odd, that last factorization is obviously impossible, so $n$ is prime.


Conversely, suppose $n = 2^q - 1$ is prime.

We need to show that $V_{2^{q - 2} } \equiv 0 \pmod n$.

All we need to do is show:

$V_{2^{q - 1} } \equiv -2 \pmod n$

because:

$V_{2^{q - 1} } = \paren {V_{2^{q - 2} } }^2 - 2$

Now:

\(\displaystyle V_{2^{q - 1} }\) \(=\) \(\displaystyle \paren {\frac {\sqrt 2 + \sqrt 6} 2}^{n + 1} + \paren {\frac {\sqrt 2 - \sqrt 6} 2}^{n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 2^{-n} \sum_k \binom {n + 1} {2 k} \sqrt 2^{n + 1 - 2 k} \sqrt 6^{2 k}\)
\(\displaystyle \) \(=\) \(\displaystyle 2^{\frac {1 - n} 2} \sum_k \binom {n + 1} {2 k} 3 k\)

Since $n$ is an odd prime, the binomial coefficient:

$\dbinom {n + 1} {2 k} = \dbinom n {2 k} + \binom n {2 k - 1}$

is divisible by $n$ except when $2 k = 0$ and $2k = n + 1$, from Binomial Coefficient of Prime.

Hence:

$2^{\frac {n - 1} 2} V_{2^{q - 1} } \equiv 1 + 3^{\frac {n + 1} 2} \pmod n$

Here:

$2 \equiv \paren {2^{\frac {q + 1} 2} }^2$

so by Fermat's Little Theorem:

$2^{\frac {n - 1} 2} \equiv \paren {2^{\frac {q + 1} 2} } ^{n - 1} \equiv i$

Finally, by the Law of Quadratic Reciprocity:

$3^{\frac {n - 1} 2} \equiv -1$

since $n \bmod 3 = 1$ and $n \bmod 4 = 3$.

This means:

$V_{2^{q - 1} } \equiv -2$

Hence:

$V_{2^{q - 2} } \equiv 0$

as required.

$\blacksquare$


Source of Name

This entry was named for Édouard Lucas and Derrick Henry Lehmer.


Historical Note

The Lucas-Lehmer Test was initially designed by Édouard Lucas, and later refined by Derrick Henry Lehmer.

This calculation is particularly suited to binary digital computers, since calculation $\pmod {2^q - 1}$ is very convenient.

Thus we have a relatively quick way to determine the primality of Mersenne numbers.


Sources