
1. Find all positive integers $a$ a $b$, such that

$$
4^{a}+4 a^{2}+4=b^{2} \text {. }
$$


Solution. The equation implies, that $b^{2}$ is even and greater than $4^{a}$, which means that $b$ is even and greater than $2^{a}$, that is $b \geqslant 2^{a}+2$, and

$$
4^{a}+4 a^{2}+4=b^{2} \geqslant\left(2^{a}+2\right)^{2}=4^{a}+4 \cdot 2^{a}+4 .
$$

which gives $a^{2} \geqslant 2^{a}$, and consequently $a \leqslant 4$.

Substituting the possible values $a=1,2,3,4$ to the original equation we find out solutions $(a, b)=(2,6)$ and $(a, b)=(4,18)$.
