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 $d\equiv 3(\mathrm{mod}4)$. Moreover, it follows from $3d=(3x-2y{)}^2+5y^2$ that $3d\equiv (3x-2y{)}^2(\mathrm{mod}5)$ and therefore $3d\equiv \pm 1(\mathrm{mod}5)$ $⟺$ $d\equiv \pm 2(\mathrm{mod}5)$ since $(d,5)=1$.

Now $d\equiv 3(\mathrm{mod}4)$ and $d\equiv \pm 2(\mathrm{mod}5)$ imply that $d\equiv 3(\mathrm{mod}20)$ or $d\equiv 7(\mathrm{mod}20)$. It is obvious that $d=3$ is not a solution, and $d=7$ gives $2^n\equiv -1(\mathrm{mod}7)$, 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 $d=23$.