5. Solve the equation $a^{b}+a+b=b^{a}$ in prime numbers.
Answer: $a=5, b=2$.

Solution. Move $b$ to the right side and group: $a\left(a^{b-1}+1\right)=b\left(b^{a-1}-1\right)$. Since numbers $a$ and $b$ are prime and, obviously, different, $a$ does not divide $b$, from which $a^{b-1}+1 \vdots b$, meaning $a^{b-1} \equiv-1(\bmod b)$. On the other hand, according to Fermat's little theorem, $a^{b-1} \equiv+1(\bmod b)$. This means that $-1 \equiv+1(\bmod b)$ or $2 \equiv 0(\bmod b)$, that is, $b=2$. We have the equation $a^{2}+a+2=2^{a}$. By direct substitution, we get that $a \neq 2$, $a \neq 3 ; a=5$ fits ( $5^{2}+5+2=32=2^{5}$ ). And for any natural $a>5$ this equation has no solutions, since increasing $a$ by one unit increases the left side by less than twice:

$\left((a+1)^{2}+(a+1)+2\right)-2\left(a^{2}+a+2\right)=\left(a^{2}+3 a+4\right)-\left(2 a^{2}+2 a+4\right)=-a^{2}+a=-a(a-1)<0$.

Remark. The problem can be solved for arbitrary natural $a$ and $b$ (Problem 5 for 11th grade).

Criteria. 2 points are given for the guessed answer.