2. Find all prime numbers $p$ for which the number $3^{p}-(p+2)^{2}$ is a prime number.
IV /2. For $p=2$, the number $3^{2}-4^{2}$ is negative. If $p$ is an odd number, $3^{p}-(p+2)^{2}$ is even; hence equal to 2. The equation $3^{p}-(p+2)^{2}=2$ has an obvious solution $p=3$. To show there are no others, we prove by induction.

More precisely: By induction, we prove that for $n>3$, $3^{n}-(n+2)^{2}>2$ or $3^{n}>n^{2}+4n+6$. For $n=4$, the estimate holds. In the inductive step, we add the inequality $2 \cdot 3^{n}>2n+5$ (valid for $n \geq 2$) to the inductive hypothesis $3^{n}>n^{2}+4n+6$ and obtain $3^{n+1}>(n+1)^{2}+4(n+1)+6$. (The proof of the inequality $2 \cdot 3^{n}>2n+5$ for $n \geq 2$ using induction is left to the reader.)