1. Solve the equation $3 \cdot 2^{x}+1=y^{2}$ in integers.
Answer: $(0 ;-2),(0 ; 2),(3 ;-5),(3 ; 5),(4 ;-7),(4 ; 7)$.

## Solution

Obviously, $x \geq 0$, otherwise the left side would be a fraction. It is also clear that $y=0$ is not a solution.

Since $y$ is in an even power, the solutions will come in pairs, for example, when $x=0$, we get two solutions $y= \pm 2$.

Now let $x>0, y>0$, then we arrive at the relation $3 \cdot 2^{x}=(y-1)(y+1)$. Here, the left side, and therefore the right side, are divisible by 3, from which there are two possible cases:

1) $y=3 z+1$. Then $2^{x}=z(3 z+2)$. The right side can only have powers of two as divisors, hence $z=2^{p}$, where $x>p \geq 0$, then $2^{x-p}=2 \cdot\left(3 \cdot 2^{p-1}+1\right)$. This equality is impossible for $p>1$, and also for $p=0$, while for $p=1$ we get $x=4, y= \pm 7$.
2) $y=3 z-1$. Then $2^{x}=z(3 z-2)$. The right side can only have powers of two as divisors, hence $z=2^{p}$, where $x \geq p \geq 0$, then $2^{x-p}=2 \cdot\left(3 \cdot 2^{p-1}-1\right)$. This equality is impossible for $p>1$. For $p=0$ we get $x=0, y= \pm 2$ (obtained earlier), for $p=1$ we get $x=3, y= \pm 5$.

## Criteria:

| All roots are found, the solution is fully justified. |  |
| :---: | :---: |
| $4-5$ | All roots are found, the solution is fully justified. |
|  | Some roots are lost with a justified solution for part of the possible values of $y$ (zero solution and/or negative $y$ are lost). |
|  | All roots are obtained with an incorrect or missing justification. |
| $\because$ | Several roots are obtained by enumeration. |