[ equations in integers $]$ [GCD and LCM. Mutual simplicity ]

Find all pairs of integers $(x, y)$, satisfying the equation $3 \cdot 2^{x}+1=y^{2}$.
The equation can be rewritten as $3 \cdot 2^{x}=(y+1)(y-1)$. The difference between the numbers $y+1$ and $y-1$ is 2, so their greatest common divisor $d$ is 1 or 2. If $d=1$, then $x=0$. If $d=2$, then either one number is $\pm 3 \cdot 2^{n}$ and the other is $\pm 2$, or one number is $\pm 3 \cdot 2$ and the other is $\pm 2^{n}$. The first case is impossible, and in the second case, $x=3$ or 4.

## Answer

$(0, \pm 2),(3, \pm 5),(4, \pm 7)$