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 $p^{12}-1$?
 A: 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 $p^{12}-1\equiv 0 \mod 13$. 
A: An approach that does not require huge computations: let $G$ be the desired number.
Suppose  that for a prime $p\neq2,5$ and positive integer $a$ we have $q^{12}\equiv 1\bmod p^a$ for every prime ending in $1$, then $\varphi(p^a)|12$ and $p$ does not end in $11$.
Proof: If $p$ ends in $11$ then $p^{12}\not\equiv 1 \bmod p^a$. If $\varphi(p)\nmid 12$ the multiplicative group of $\mathbb Z_{p^a}$ is isomorphic to $\mathbb Z_{\varphi(p^a)}$, so if $\varphi(p^a)\nmid 12$ there is a congruence class $w \bmod p^{a}$ so that $w^a\not\cong 1 \bmod p^a$. Take $n$ so that $n\equiv w \bmod p^a$ and $n\equiv 1 \bmod 10$ (possible by CRT). By dirichlet's theorem there exists a prime $r$ with $r\equiv n \bmod 10p$. Notice that $p^a \nmid r^a-1$.
On the other hand, if $\varphi(p)|12$ and $q$ does not end in $1$ then $q^a\equiv 1 \bmod p^a$ for every number $q$ that is not a multiple of $q$. And clearly no prime ends in $1$ and is a multiple of $q$.
It is easy to see that $\varphi(p^a)|12$ only for $3^1,3^2,7,13$ (where $p\neq 2,3)$.
So $G=2^a\times3^3\times5^b\times7\times 13$
all that remains is to find $a$ and $b$. It is clear that $a\geq 4$ and $5\geq1$.
Since $11\equiv 1 \bmod 5$ the order of $11\bmod 25$ can be only $1$ or $5$,it is clearly not $1$, therefore $5^{12}\not\equiv 1 \bmod 25$, we conclude $b=1$ (clearly $b\geq 1$ since $\varphi(5)|12$)
On the other hand the order of $11\bmod 32$ must be a power of $2$, if it divides $12$ then we must have $11^4\equiv 1 \bmod 32$, we have $11^4\equiv 11^211^2\equiv -7\times-7\equiv 49\not\equiv 1 \bmod 32$. we conclude $b\leq 4$, clearly $b\geq 4$ as the multiplicative group of $\mathbb Z_{2^4}$ is congruent to $\mathbb Z_{2^2}\times \mathbb Z_2$ and every element in this group has an order that divides $12$.
We conclude $G=2^4\times3^2\times5\times 7\times 13$
