More on Fermat witnesses and liars

In my previous post I stated, without proof, the following theorem:

Theorem: if n is composite and there exists at least one Fermat witness for n, then at least half of the numbers relatively prime to n are Fermat witnesses.

Were you able to prove this? Here’s one way to do it.

Suppose n is composite and a is a Fermat witness for n—that is, a^{n-1} \not\equiv 1 \pmod n. Let b be some Fermat liar for n, that is, b is relatively prime to n but b^{n-1} \equiv 1 \pmod n. (How do we know there exists a Fermat liar for n, you ask? Well, we don’t, but if there aren’t any then the theorem is obviously true.) Then I claim that ab is also a Fermat witness for n:

(ab)^{n-1} = a^{n-1} b^{n-1} \equiv a^{n-1} \cdot 1 \not \equiv 1 \pmod n

So for every Fermat liar b there is a corresponding Fermat witness ab. The only thing we might worry about is that some of these Fermat witnesses might not be distinct. But in fact, since a is relatively prime to n, multiplication by a modulo n is invertible, so ab_1 and ab_2 must be distinct modulo n whenever b_1 and b_2 are. This proves that there are at least as many Fermat witnesses as there are liars; hence at least half are witnesses.

This means that, as long as there is at least one Fermat witness for n, each a we pick has a probability of at least 1/2 of revealing that n is composite. So if we pick k different values for a and all of them yield a^{n-1} \equiv 1 \pmod n, we can conclude that there is only a 1/2^k chance that a is composite. That’s pretty good, and it’s nice that the probability is independent of n.

However, there’s one big flaw, which relates to my third question from the previous post: can there be composite values of n with no Fermat witnesses? More in my next post!

About Brent

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