Four formats for Fermat: correction!

In my previous post I explained three variants on Fermat’s Little Theorem, as well as a fourth, slightly more general variant, which it turns out is often called Euler’s Totient Theorem. Here’s what I said:

If n \geq 1 and a is any integer, then a^{\varphi(n)} \equiv 1 \pmod n.

However, this is wrong! We can easily find a counterexample. For example, let n = 6 and a = 3. Then \varphi(6) = 2, since there are only two numbers less than 6 that are relatively prime to it (namely, 1 and 5; each of 2, 3, and 4 shares a common factor with 6). But then a^{\varphi(n)} = 3^2 = 9 which has a remainder of 3, not 1, when divided by 6.

What’s wrong? It turns out I missed a small but important restriction: this is only true when a is relatively prime to n. Note this is stronger than saying a isn’t divisible by n; it says they cannot share any common factors at all. For example, 9 is not divisible by 6, but they share 3 as a common factor, so this theorem does not apply (and indeed, 9^2 \equiv 3 \pmod 6, not 1).

For completeness, here’s a corrected statement of the theorem (I have also fixed the previous post):

If n \geq 1 and a is any integer relatively prime to n, then a^{\varphi(n)} \equiv 1 \pmod n.

So what’s an example where this theorem does work? Let’s keep n = 6 but try a = 5 instead: now a^{\varphi(6)} = 5^2 = 25 \equiv 1 \pmod 6.

In my next post I’ll present a proof of Fermat’s Little Theorem.

About Brent

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