# Primitive roots in arithmetic progression

Let $$a$$ be a primitive root modulo odd prime. Show that in an arithmetic progression $$a+kp$$, where $$k = 0,1,\dots,p-1$$ there is exactly one number that is NOT a primitive root modulo $$p^2$$.

It is obvious that all of these are congruent to $$a$$ modulo $$p$$ so they are all primitive roots modulo $$p$$. To show that all but one are primitive roots modulo $$p^2$$ as well we would need to see that $$(a+kp)^{p-1} \equiv 1$$ (mod $$p^2$$) for exactly one value of $$k$$. I tried to do so using the Newton binomial but right now have failed to do so? Any suggestions?

• $(\Bbb{Z/p^2 Z})^\times$ is a group with $p(p-1)$ elements, thus it contains an element $b=c+dp$ of order $p$, reducing modulo $p$ means $c=1$. Whence $(1+dp)^m a^n$ is of order $\frac{p}{gcd(m,p)}\frac{p-1}{gcd(n,p-1)}$. Note you can take $da \equiv 1 \bmod p, (1+d p)^m = 1+dmp\bmod p^2,(1+dp)^m a= a+damp=a+mp\bmod p^2$. – reuns Apr 29 at 8:14
• math.stackexchange.com/questions/227199/… – lab bhattacharjee Apr 29 at 9:43