2017 AIME II Problems/Problem 6

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Problem

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

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 $168+27=\boxed{195}$.

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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