Find all primes that satisfy congruency $100^p\equiv1\mod p$
I've tried reducing it to the fact that $100^p=(10^p)^2$ so then $10^p \equiv 1 \mod p$ or $-1 \mod p$.
I've also attempted writing this as $100^p-np=1$, so that leads me to $\gcd(100^p,n)=1$ but I don't see where to go from here.
Would appreciate some help with this! I only know a little bit about number theory so preferably an answer using more basic stuff (Like Fermat's little theorem, linear diophantine equations, gcd's)
Edit after reading comments: Following from Fermat's little theorem, since we know $100^p\equiv100\mod p$ for all prime p, but $100^p\equiv1\mod p$ only if $100-np=1$ so $np=99$ so $p=11,3$ are the only solutions? Is this correct?