2017 AIME II Problems/Problem 6

$\textbf{Problem 6}$ Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

$\textbf{Problem 6 Solution}$ Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$ . The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$ . Rearranging, $s^2-(2n+85)^2=843$ . By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$ . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$ . The two values of $n$ that satisfy one of the equations are $168$ and $27$ . Summing these together, the answer is $168+27=\boxed{195}$ .

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2017 AIME II Problems/Problem 6