A Probabilistic Primality Test

Recall,

Theorem 3.1   A natural number $p$ is prime if and only if for every $a\not\equiv 0\pmod{p}$,

$$a^{p-1}\equiv 1\pmod{p}.$$

Thus if $p\in\mathbb{N}$ and, e.g., $2^{p-1}\not\equiv 1\pmod{p}$, then we have proved that $p$ is not prime. If, however, $a^{p-1}\equiv 1\pmod{p}$ for a couple of $a$, then it is ``highly likely'' that $p$ is prime. I will not analyze this probability here, but we might later in this course.

Example 3.2   Let $p=323$. Is $p$ prime? Let's compute $2^{322}$ modulo $323$. Making a table as above, we have

$i$	$m$	$\varepsilon _i$	$2^{2^i}$ mod 323
0	322	0	2
1	161	1	4
2	80	0	16
3	40	0	256
4	20	0	290
5	10	0	120
6	5	1	188
7	2	0	137
8	1	1	35

Thus

$$2^{322} \equiv 4\cdot 188\cdot 35 \equiv 157\pmod{323},$$

so $323$ is not prime. In fact, $323 = 17\cdot 19$.

It's possible to prove that a large number is composite, but yet be unable to (easily) find a factorization! For example if

$$n = 95468093486093450983409583409850934850938459083,$$

then $2^{n-1}\not\equiv 1\pmod{n}$, so $n$ is composite. This is something one could verify in a reasonable amount of time by hand. (Though finding a factorization by hand would be very difficult!)