Prove g is a primitive root of p

Question

I know of the method to calculate $g^k$ for every $k < p$ and check if it ever is congruent to $1$.

But can I prove the same property with less work by taking the factors of $(p-1)$ and only raising g to those powers?

For Example: $g = 2$, $p = 7$ we can find $p-1 = 6$ and the factors of 6 are $1,2,3,6$ and can we only check these factors. $2^1 = 2$, $2^2 = 4$, $2^3 = 1$, $2^6 = 1$. Here shows that $2$ is not a primitive root of $7$ since it should have an order = $6$ but has an order = $3$.

Yes. But do you have a general way to find them without finding them one at a time? (In any case you can stop halfway.) — Ethan Bolker, Oct 23 '22 at 18:27
@EthanBolker I know of one but I was more looking for how to prove a specific number is a primitive root mod p. — Anonymous98737943, Oct 23 '22 at 18:35
Yes there is a faster way to compute only some powers: use the exponentiation by squaring algorithm. For instance, to compute $g^{20}$, you only have to compute $g^2$, $g^4$, $g^8$, $g^{16}$ and $g^{20} = g^{16} g^4$. — Tomek Czajka, Oct 23 '22 at 18:40

score 2 · Answer 1 · answered Oct 23 '22 at 18:36

2

You can do even better: only consider prime factors of $p-1$.

$g \not\equiv 0$ is a primitive root mod $p$ if for every prime factor $q | p - 1$: $$g^\frac{p-1}{q} \not\equiv 1 \pmod{p}$$

That's because every proper divisor of $p-1$ is also a divisor of $\frac{p-1}{q}$ for some $q$.

answered Oct 23 '22 at 18:36

Tomek Czajka

361

So for my example I would only have to consider 2,3 for modulo 7? – Anonymous98737943 Oct 23 '22 at 18:41
Yes. Only check $2^2$ and $2^3$ for your example. – Tomek Czajka Oct 23 '22 at 18:41
Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 24 '22 at 17:47

Prove g is a primitive root of p

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