The integers $a$ and $b$ have the property that the expression

$$
\frac{2 n^{3}+3 n^{2}+a n+b}{n^{2}+1}
$$

is an integer for every integer $n$. What is the value of the expression above when $n=4$ ?
By rearranging, we get

$$
\begin{aligned}
\frac{2 n^{3}+3 n^{2}+a n+b}{n^{2}+1} & =\frac{2 n^{3}+2 n+3 n^{2}+3+(a-2) n+(b-3)}{n^{2}+1} \\
& =\frac{2 n\left(n^{2}+1\right)+3\left(n^{2}+1\right)+(a-2) n+(b-3)}{n^{2}+1} \\
& =2 n+3+\frac{(a-2) n+(b-3)}{n^{2}+1}
\end{aligned}
$$

The quantity $2 n+3$ is an integer if $n$ is an integer, so the given expression in $n$ is an integer exactly when $\frac{(a-2) n+(b-3)}{n^{2}+1}$ is an integer.

Suppose that $a-2 \neq 0$ and consider the functions $f(x)=x^{2}+1, g(x)=(a-2) x+b-3$, and $h(x)=f(x)-g(x)=x^{2}-(a-2) x+4-b$.

The graph of $h(x)$ is a parabola with a positive coefficient of $x^{2}$, which means that there are finitely many integers $n$ with the property that $h(n) \leq 0$ (there could be no such integers). Since $h(n)=f(n)-g(n)$, we have shown that there are only finitely many integers $n$ for which $g(n)0$.

Since infinitely many integers satisfy $0<g(n)$ and only finitely many integers fail $g(n)<f(n)$, there must exist an integer $n$ with the property that $0<g(n)<f(n)$.

This means there exists an integer $n$ for which

$$
0<\frac{g(n)}{f(n)}<1
$$

but this means

$$
\frac{g(n)}{f(n)}=\frac{(a-2) n+b-3}{n^{2}+1}
$$

is not an integer. Therefore, we have shown that if $a-2 \neq 0$, then there is an integer $n$ for which the expression in the problem is not an integer, so we conclude that $a-2=0$ or $a-2$. We now have that the expression in the question is equivalent to

$$
2 n+3+\frac{b-3}{n^{2}+1}
$$

which is an integer exactly when $\frac{b-3}{n^{2}+1}$ is an integer.

If $b-3$ is positive, then $n$ can be chosen so that $n^{2}+1$ is larger than $b-3$, which would mean that the positive expression $\frac{b-3}{n^{2}+1}$ is not an integer.

Similarly, if $b-3$ is negative, then there are integers $n$ for which $\frac{b-3}{n^{2}+1}$ is not an integer.

This shows that if $b-3 \neq 0$, then there are integers $n$ for which the expression in the question is not an integer.

We can conclude that $b-3=0$, or $b=3$.

Therefore, we have that $a=2$ and $b=3$, so the expression given in the problem is equivalent to

$$
2 n+3+\frac{(2-2) n+(3-3)}{n^{2}+1}=2 n+3
$$

Thus, the answer to the question is $2(4)+3=11$.

ANSWER: 11