2. Find all integer solutions to the equation $a^{b}=a b+2$.

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However, it seems there's a redundancy in my response. The text has already been translated. Here is the translation:

2. Find all integer solutions to the equation $a^{b}=a b+2$.
IV/2. If $b1$. For $a=1$, we get the equation $1=b+2$ and from this $b=-1$. For $a=-1$, we get the equation $(-1)^{b}=-b+2$, which has no solution: since $b>0$, the number $a$ divides $a^{b}$. Therefore, $a$ also divides the number 2, since $a^{b}-a b=2$. Thus, $a$ can be equal to $-2, -1, 1$ or 2. Let's consider each of these cases separately.

If $a=-2$, we have $(-2)^{b}=-2b+2$ or $(-2)^{b-1}=(b-1)$. This equation is not satisfied by any natural number $b$. If the number $b$ satisfies the condition, it is odd; thus $b=2s+1$. The equation then becomes $4^{s}=s$, which is not solvable, as for any real number $s$, $4^{s}>s$.

If $a=-1$, we have $(-1)^{b}=-b+2$. This means that $|-b+2|=1$, which gives us $b=1$ or $b=3$, but only $b=3$ satisfies the equation $(-1)^{b}=-b+2$.

If $a=1$, we get the equation $1=b+2$, which has no solution for $b>0$.

If $a=2$, we have $2^{b}=2(b+1)$ or $2^{b-1}=b+1$. This equation is only satisfied by $b=3$. For $b>3$, $2^{b-1}>b+1$, which we can verify by induction. (For $b=4$, we have $2^{4-1}=8>4+1=5$. In the inductive step, $2^{(b+1)-1}=2 \cdot 2^{b-1}>2(b+1)>(b+1)+2$.)

The pairs $(a, b)$ that solve the equation are: $(2,3)$, $(1,-1)$, and $(-1,3)$.