2. Find all triples of natural numbers $a, b$, and $c$, for which the numbers $a^{2}-23$ and $b^{2}-23$ are prime and $\left(a^{2}-23\right)\left(b^{2}-23\right)=c^{2}-23$.
Answer: $(5,6,7)$ and $(6,5,7)$.

Solution. Note that $a, b \geqslant 5$. If $a=b=5$, then $c^{2}-23=\left(a^{2}-23\right)\left(b^{2}-23\right)=2 \cdot 2=4$, which is impossible. Therefore, we can assume that $a \geqslant b$ and $a \geqslant 6$. Then

$$
c^{2}-23=\left(a^{2}-23\right)\left(b^{2}-23\right) \leqslant\left(a^{2}-23\right)^{2}
$$

On the other hand, $(c+a)(c-a)=\left(c^{2}-23\right)-\left(a^{2}-23\right)$ is divisible by $a^{2}-23$. If $c-a$ is divisible by $a^{2}-23$, then $c-a \geqslant a^{2}-23$ and $c+a>c-a \geqslant a^{2}-23$, and thus

$$
\left(a^{2}-23\right)^{2}<(c+a)(c-a)=\left(c^{2}-23\right)-\left(a^{2}-23\right) \leqslant\left(a^{2}-23\right)^{2}
$$

which is impossible. Therefore, $c+a$ is divisible by $a^{2}-23$. If $c+a \geqslant 2\left(a^{2}-23\right)$, then

$$
c \geqslant 2 a^{2}-a-46 \geqslant a^{2}+6 a-a-46=a^{2}+5 a-46 \geqslant a^{2}+5 \cdot 6-46=a^{2}-16
$$

and thus

$$
a^{4}-46 a^{2}+529=\left(a^{2}-23\right)^{2} \geqslant c^{2}-23 \geqslant\left(a^{2}-16\right)^{2}-23=a^{4}-32 a^{2}+233
$$

Therefore, $296 \geqslant 14 a^{2} \geqslant 14 \cdot 6^{2}=504$, which is impossible. Therefore, $c+a=a^{2}-23$ and $c-a=b^{2}-24$. Since $a \geqslant 6$, the prime number $a^{2}-23$ is at least 13 and thus is odd. Therefore, $c=a^{2}-a-23$ is also odd, and thus the number $c^{2}-23$ is even. But it is the product of two prime numbers $a^{2}-23$ and $b^{2}-23$, so $b^{2}-23=2$ and $b=5$. Thus, $c-a=b^{2}-24=1$, from which $a+1=c=a^{2}-a-23$ and thus $a=6$. It is easy to see that the triple $a=6, b=5$ and $c=7$ satisfies the condition of the problem.