Lucas-Lehmer Test

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:

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:

Similarly:

In general:

Taking $k = p$, we get:

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

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:

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 $L_{q - 2} = 0$:

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

We have:

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

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:

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.

$\blacksquare$

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Source of Name

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

Historical Note

The was initially designed by François Édouard Anatole 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

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2^{521} - 1$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2^{521} - 1$