67th NMO Shortlisted Problems

Let $2n^2+27n+91=k^2$ for some $k\in \mathbb{N}$. Rearranging, we get:

$$2n^2+27n+91-k^2=0$$

This is a quadratic in $n$:

$$2n^2+27n+(91-k^2)=0$$

The discriminant must be a perfect square for $n$ to be integer:

$$\Delta =27^2-4\cdot 2\cdot (91-k^2)=729-8(91-k^2)=729-728+8k^2=1+8k^2$$

So $1+8k^2$ must be a perfect square. Let $1+8k^2=m^2$ for some $m\in \mathbb{N}$.

Then $m^2-8k^2=1$, which is a Pell equation.

The fundamental solution is $(m_1,k_1)=(3,1)$, since $3^2-8\cdot 1^2=1$.

All solutions are given by:

$$(m+k\sqrt{8})=(3+\sqrt{8}{)}^t$$ for $t\in \mathbb{N}$.

For each such $k$, we can solve for $n$:

$$n=\frac{-27\pm \sqrt{1+8k^2}}{4}$$

Since $\sqrt{1+8k^2}=m$, and $m$ is odd (since $m^2\equiv 1(\mathrm{mod}8)$), $-27+m$ is even, so $n$ is integer.

Thus, for each solution $(m,k)$ to the Pell equation, we get an integer $n$ such that $2n^2+27n+91$ is a perfect square.

Therefore, there are infinitely many such $n\in \mathbb{N}$.