MaBloWriMo 6: The Proof Begins

Today we’re going to start in on proving the Lucas-Lehmer test. Yesterday we saw how, given a Mersenne number M_n = 2^n - 1, we can define a sequence of integers s_0, s_1, \dots, with the claim that s_{n-2} = 0 if and only if M_n is prime. We’re going to start by focusing on the proof of the “only if” direction, that is, if s_{n-2} = 0, then M_n is prime. Once we prove this, we’ll know for sure that the Lucas-Lehmer test produces no false positives: if it says M_n is prime, then it definitely is. Proving the converse—if M_n is prime, then s_{n-2} = 0—would mean that the test doesn’t miss any primes. That proof is a bit harder; we’ll see how far we get this month!

I’ll mostly be following Bruce (1993), which is in turn based on Rosen (1988). As is my habit, I’ll expand quite a bit, filling in missing details and exploring interesting tangents as we go.

And so it begins

Define \omega = 2 + \sqrt{3} and \overline{\omega} = 2 - \sqrt{3}. We can easily check that \omega \overline\omega = 1:

\omega \overline\omega = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 2\sqrt{3} + 2\sqrt{3} - 3 = 1.

So what on earth do \omega and \overline{\omega} have to do with the Lucas-Lehmer test? Well, here’s a challenge for you: prove that

s_m = \omega^{2^m} + \overline{\omega}^{2^m} \pmod{M_n}.

where s_m are the numbers we defined yesterday. (Hint: use induction on m.) Solution to come tomorrow!


Bruce, J. W. 1993. “A Really Trivial Proof of the Lucas-Lehmer Test.” The American Mathematical Monthly 100 (4). Mathematical Association of America: 370–71.

Rosen, Michael I. 1988. “A Proof of the Lucas-Lehmer Test.” Am. Math. Monthly 95 (9). Washington, DC, USA: Mathematical Association of America: 855–56. doi:10.2307/2322904.


About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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One Response to MaBloWriMo 6: The Proof Begins

  1. Pingback: MaBloWriMo 7: s via omega | The Math Less Traveled

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