2. Find all integer solutions of the equation $3 \cdot 2^{x}+1=y^{2}$.
Answer. (0,2), $(3,5),(4,7)$.

Solution. Note that $x \geq 0$ (otherwise $y$ is not an integer). If $x=0$, then $y=2$. Let $x$ be a natural number, then $y$ is odd, denote $y=2k+1$. We get $3 \cdot 2^{x}+1=(2k+1)^{2}$, or $3 \cdot 2^{x}=4k^{2}+4k$, from which $3 \cdot 2^{x-2}=k(k+1)$. Since $k$ and $k+1$ are coprime, the possible cases are $k=3, k+1=2^{x-2}$ or $k+1=3, k=2^{x-2}$. The first case gives the solution $x=4, y=7$, in the second case $x=3, y=5$.

Comment. It is shown that for $x$ natural, $y$ is odd - 1 point. If the solutions of the equation are found by trial and error, and it is not proven that there are no others - 1 point for each found solution.