2. Determine all pairs of positive integers $(m, n)$ for which

$$
3 \cdot 2^{n}+1=m^{2} .
$$
2. Determine all pairs of positive integers $(m, n)$ for which

$$
3 \cdot 2^{n}+1=m^{2} .
$$

Solution The equation can be rewritten as $3 \cdot 2^{n} = (m-1)(m+1)$. Since $n>0$, $3 \cdot 2^{n}$ is even, so either $m-1$ or $m+1$ is even, and thus both are even. Because the difference between $m-1$ and $m+1$ is only 2, they cannot both have more than one factor of 2, so one of the two factors $m-1$ and $m+1$ must have exactly one factor of 2. This single factor may or may not also contain the factor 3 and is therefore equal to 2 or 6; the other factor is 2 higher or 2 lower.
If one factor is 2, then the other factor must be 0 or 4, which in both cases does not yield a solution (since the other factor must be of the form $3 \cdot 2^{n-1}$). If one factor is 6, then the other is 4 or 8, and both possibilities lead to a solution (since the other factor must now be of the form $2^{n-1}$). We conclude that there are exactly two solutions: $(m, n)=(5,3)$ and $(m, n)=(7,4)$.