Show that if ab has a primitive root with $(a,b) = 1$, then $a<3$ or $b<3$ Show that if $ab$ has a primitive root with $\gcd\left(a,b\right) = 1$, then $a<3$ or $b<3$
I have no idea how to start this question at all...
One is that I do not see how 3 is related to this question also how the concept primitive root can be related to the values of a and b...
Thanks in advance
 A: If this is within the scope of permitted answer, it's probably worth thinking about the Carmichael function $\lambda$. This give the minimum order of any $a$ coprime to the base $n$, and differs from Euler's totient function $\phi$ in two respects: higher powers of $2$, from $8$ onwards, have $\lambda(n)= \phi(n)/2$, and more importantly, $\lambda(p_1^{k_1}p_2^{k_2}) = \gcd(\lambda(p_2^{k_2}) ,\lambda(p_2^{k_2}) )$. However for odd prime powers, $\lambda(n)= \phi(n)$.
Now since we're talking about some value $n=ab$ that is clearly not prime or a prime power, and since both $\phi(n) $ and $\lambda(n)$ are even for $n\ge 3$, the $\gcd$ combination would ensure that the highest multiplicative order of any qualifying number, $\lambda(n)$, is less than the number of numbers coprime to $n$, $\phi(n)$, unless one of the factors of $n$ is exactly $2$ and the other is an odd prime power. Since the existence of a primitive root requires that the order of that root is equal to $\phi(n)$, we can be sure that the described exception is the case, and one of $a,b$ is exactly $2$.

Edit to add: I see that it is also possible that $a=1$ or $b=1$ - I automatically excluded that case, but the terms of the question do not - in which case again one of $a,b<3$.
