Shortest way finding a primitive root of $15$

Question

Note that Euler phi function $\phi(15)=8$. Note that $\{ 2,\ 4,\ 7,\ 8,\ 11,\ 13,\ 14 \}$ is the set of relative numbers to $ 15$. And $$ 2^4\equiv 1\ (15)$$ so that since $4<\phi(15)$, $2$ is not primitive. So we completed by testing six times more, and we concluded that there exists no primitive.

Here I have a question : Is there more shorter proof ?

The "Finding Primitive Roots" section of the Primitive Roots page of Wikipedia has a relevant discussion. Also see this earlier question. — Semiclassical, Jul 27 '14 at 04:19

score 1 · Accepted Answer · answered Jul 27 '14 at 04:52

1

The shortest way is to quote the general theorem about which numbers have a primitive root.

But for $15$, note that $\varphi(3)=2$, and $\varphi(5)=4$. The lcm of these is $4$, so if $a$ is relatively prime to $15$, we have $a^4\equiv 1\pmod{15}$. Since $\varphi(15)=8$, there is no primitive root.

answered Jul 27 '14 at 04:52

André Nicolas

507,029

$(a^4-1)(a^4+1)\equiv 0\ (15)$ and $a^4 \equiv 1\ (5)$. So we conclude. Thank you. – HK Lee Jul 27 '14 at 05:00
You are welcome. For more details, you may want to look at the Carmichael function. – André Nicolas Jul 27 '14 at 05:01

Shortest way finding a primitive root of $15$

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