2. Find all primes $p$ for which the numbers $p+7$ and $p^{2}+7$ are double the squares of natural numbers.
Answer: $p=11$

First solution. Let $p+7=2 x^{2}$ and $p^{2}+7=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Since $p-$ is odd, $p \geqslant 3$ and $2 p^{2}>p^{2}+7$. Therefore, $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note that $\frac{y^{2}}{x^{2}}=\frac{p^{2}+7}{p+7}>\frac{p^{2}-49}{p+7}=p-7$. Then if $p \geqslant 17$, $y^{2}>10 x^{2}$, from which $y>3 x$ and, thus, $2(y-x)>y+x \geqslant p$. Therefore, $2(y-x)(y+x)>p^{2}>p(p+1)$, which is also impossible. Thus, it remains to consider the cases $p=3, p=5, p=7, p=11$ and $p=13$. In the last, as well as in the first three of them, $p+7$ is not a doubled square, while $p=11$ fits.