2. Solve the equation $2^{2 p}-2^{p}+1=q$, where $p$ and $q$ are prime numbers.

Answer: $p=2, q=13$.
Solution. Let $p=2$, then $q=13$.

Now let $p$ be an odd prime. Then the left side of the equation is divisible by 3 (since 2 to an odd power gives a remainder of 2 when divided by 3, and to an even power - 1), so $q=3$, from which $\left(2^{p}-2\right)\left(2^{p}+1\right)=0$, i.e., there are no solutions.