Fall Mathematical Competition

Let $d = 3x^2 - 4xy + 3y^2$ for some integers $x$ and $y$ and suppose that $d$ divides $2^n + 15$ for some positive integer $n$. Obviously $d$ is odd and this implies that $x$ and $y$ have different parity. Then we have $d \equiv 3 \pmod{4}$. Moreover, it follows from $3d = (3x - 2y)^2 + 5y^2$ that $3d \equiv (3x - 2y)^2 \pmod{5}$ and therefore $3d \equiv \pm 1 \pmod{5}$ $\Longleftrightarrow$ $d \equiv \pm 2 \pmod{5}$ since $(d, 5) = 1$.

Now $d \equiv 3 \pmod{4}$ and $d \equiv \pm 2 \pmod{5}$ imply that $d \equiv 3 \pmod{20}$ or $d \equiv 7 \pmod{20}$. It is obvious that $d = 3$ is not a solution, and $d = 7$ gives $2^n \equiv -1 \pmod{7}$, which is also impossible. The next possibility $d = 23$ satisfies the conditions of the problem for $n = 3$ and $x = 2, y = -1$. Therefore the required number is $d = 23$.