PROBLEM 2. a) Show that $x^{2}+x-2=(x-1)(x+2)$, for any $x \in \mathbb{R}$;

b) Determine the natural numbers $a, b$ and the prime number $p$, knowing that $a^{2}+a=p^{2^{b}}+2$.
## PROBLEM 2.

$(2 p)$

a) It is shown that $x^{2}+x-2=(x-1)(x+2)$

$(1 p) \quad$ b) Since the number $a^{2}+a=a(a+1)$ is even, it follows that the number $p^{2^{b}}+2$ is also even. Therefore, $p=2$, and the relation from the statement becomes: $a^{2}+a-2=2^{2^{b}} \Longleftrightarrow(a-1)(a+2)=2^{2^{b}}$

(1p) Since the right-hand side is even, and the numbers $a-1$ and $a+2$ have different parities, the only possibilities are: $\left\{\begin{array}{l}a-1=1 \\ a+2=2^{2^{b}}\end{array}\right.$ or $\left\{\begin{array}{l}a-1=2^{2^{b}} \\ a+2=1\end{array}\right.$

(1p) For $a+2=1$, there are no solutions

(2p) For $\left\{\begin{array}{l}a-1=1 \\ a+2=2^{2^{b}}\end{array}\right.$, we obtain $\left\{\begin{array}{l}a=2 \\ b=1\end{array}\right.$

The solution is $(p, a, b)=(2,2,1)$