The greatest common divisors of all numbers that are one less than the twelfth power of a prime > What is the greatest common divisor of all numbers that are one less than the twelfth power of a prime with a units digit of one? By Dirichlet's theorem, there exist infinitely many primes with a units digit of $1$. How can we find the greatest common divisor using the fact that our numbers are of the form $p^{12}-1$?

It is not more than $2^4 \times 3^2 \times 5 \times 7 \times 13$. To see this, take the $\gcd(61^{12}-1,71^{12}-1)$.

To show that this is indeed the $\gcd$, note that the fact that the prime's unit digit is a one implies that it is not any of $2, 3, 5, 7$ or $13$, so its coprime with those integers.

Note that $a^4 \equiv 1 \mod 16$ for all $a$ comprime with $16$. This can be proved by case inspection. So $$p^{12}-1=(p^3)^4 -1 \equiv 0 \mod 16$$

The Euler-Fermat theorem gives the remaining cases:

* $\varphi(5)=4$, so $p^4\equiv1\mod 5$, so $p^{12}-1\equiv 0 \mod 5$.

* $\varphi(7)=6$, so $p^6\equiv1\mod 7$, so $p^{12}-1\equiv 0 \mod 7$.

* $\varphi(9)=6$, so $p^6\equiv1\mod 9$, so $p^{12}-1\equiv 0 \mod 9$.

* $\varphi(13)=12$, so $p^{12}\equiv1\mod 13$, so $p^{12}-1\equiv 0 \mod 13$.