Tell whether $\dfrac{10^{91}-1}{9}$ is prime or not? I really have no idea how to start. The only theorem considering prime numbers I know of is Fermat's little theorem and maybe its related with binomial theorem.
Any help will be appreciated.

Using Why $a^n - b^n$ is divisible by $a-b$?,
$$a^{91}-1=(a^{13})^7-1^7$$ is divisible by $a^{13}-1$ and by similarly by $a^7-1$
Again, using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,
$$(a^7-1,a^{13}-1)=a^{(7,13)}-1=a-1$$
Hence, $\displaystyle\frac{a^{13}-1}{a-1}, \frac{a^7-1}{a-1}$ are co-prime primes and divisors of $\displaystyle\frac{a^{91}-1}{a-1}$
