. Find all integers $n$ such that there exist two twin primes $p$ and $q$ (i.e., such that $q=p+2$) such that $2^{n}+p$ and $2^{n}+q$ are also two twin primes.
(arithmetic). It is about determining the integers $n$ for which there exists a prime number $p$ such that $p+2, p+2^{n}$, and $p+2+2^{n}$ are all prime. If we look at these four numbers modulo 3, we get:

if $n$ is even, $p, p+2, p+1$, and $p$

if $n$ is odd, $p, p+2, p+2$, and $p+1$

In all cases, there is among these numbers a multiple of 3, which cannot be prime if it is different from 3. We deduce that necessarily $p=3$, and also that $n$ must be odd: we write $n=2 m+1$.

It is now a matter of finding the integers $m$ such that $2 \times 4^{m}+3$ and $2 \times 4^{m}+5$ are prime. Modulo 5, the powers of 4 are alternately 1 and -1. Thus, if $m$ is even, $2 \times 4^{m}+3$ is divisible by 5 and therefore cannot be prime, unless it equals 5. We deduce that either $m=0$, or $m$ is odd. The first possibility gives a solution which is $n=1$. We now assume $m$ is odd and write $m=2 s+1$, i.e., $n=4 s+3$. We then have $2^{n}=8 \times 16^{s}$. This quantity is congruent to $2^{s}$ modulo 7 and by studying the sequence of powers as before, we realize that $s$ must be a multiple of 3. Thus, we write $s=3 t$, i.e., $n=12 t+3$. Finally, we reason modulo 13: Fermat's little theorem shows that we then have $2^{n} \equiv 8 \pmod{13}$ and therefore that 13 divides $2^{n}+5$; thus it cannot be prime unless it equals 13 (which corresponds to $n=3$).

In the end, the only solutions are $n=1$ and $n=3$.