MaBloWriMo 5: The Lucas-Lehmer Test

We now know that $M_n = 2^n - 1$ can only be prime when $n$ is prime; but even when $n$ is prime, sometimes $M_n$ is prime and sometimes it isn’t. The Lucas-Lehmer test is a way to tell us whether $M_n$ is prime, for any prime $n > 2$ .

The test itself is actually straightforward, so I will just explain it quickly (if you want a more detailed explanation, along with some Haskell code, see my post from three years ago.

Define $s_0 = 4$ .
For each $k > 0$ , define $s_k = (s_{k-1}^2 - 2) \bmod {M_n}$ .
Then $M_n$ is prime if and only if $s_{n-2} = 0$ .

That is, start with $4$ , and at each step, square the current number, subtract two, and take the remainder when divided by $M_n$ . Do this $n-2$ times. $M_n$ is prime if you end with zero, and composite if you don’t.

Let’s do two small examples, one when $M_n$ is prime and one when it isn’t.

For $M_7 = 127$ , we get

$\begin{array}{c|cccccc} k & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline s_k & 4 & 14 & 67 & 42 & 111 & 0 \end{array}$

so $M_7$ is prime. For $M_{11} = 2047$ , we get

$\begin{array}{c|cccccccccc} k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline s_k & 4 & 14 & 194 & 788 & 701 & 119 & 1877 & 240 & 282 & 1736 \end{array}$

so $M_{11}$ is not prime (though note this doesn’t really help us factor it!).

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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3 Responses to MaBloWriMo 5: The Lucas-Lehmer Test

Pingback: MaBloWriMo 6: The Proof Begins | The Math Less Traveled
janhrcek says:

November 8, 2015 at 2:33 am

The first sentence should have $M_n = 2^n – 1$ I believe. Anyway, this series of blog posts seems really nice. Thanks for that!
- Brent says:
  
  November 9, 2015 at 1:14 pm
  
  Thanks, fixed!

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