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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 ?

HK Lee
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    The ["Finding Primitive Roots"](http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots) section of the Primitive Roots page of Wikipedia has a relevant discussion. Also see this earlier [question](http://math.stackexchange.com/questions/124408/finding-a-primitive-root-of-a-prime-number). – Semiclassical Jul 27 '14 at 04:19

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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.

André Nicolas
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