1. Determine all pairs of positive integers \( a \) and \( b \) for which

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

(Martin Panák)
Solution. From the equation, it follows that $b^{2}$ is an even number greater than $4^{a}$, so $b$ is an even number greater than the even number $2^{a}$. Therefore, it must hold that $b \geqq 2^{a}+2$, from which

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

By comparing the extreme expressions, we get $a^{2} \geqq 2^{a}$, which means that $a \leqq 4$. We can prove by induction that the opposite inequality $a^{2}<2^{a}$ holds for every integer $a \geqq 5$. For $a=5$, this is true $(25<32)$; if $a^{2}<2^{a}$ holds for some $a \geqq 5$, then multiplying by two gives $2 a^{2}<2^{a+1}$, so the desired inequality $(a+1)^{2}<2^{a+1}$ is a consequence of the inequality $(a+1)^{2}<2 a^{2}$, which is obvious, as it is equivalent to the inequality $1<a(a-2)$, which trivially holds for any $a \geqq 5$. This completes the proof by induction.

We have shown that in every sought pair $(a, b)$, it must hold that $a \leqq 4$. By successively substituting the values $a=1,2,3,4$ into the equation $4^{a}+4 a^{2}+4=b^{2}$, we find that the problem has exactly two solutions, namely $(a, b)=(2,6)$ and $(a, b)=(4,18)$.