2017 AIME II Problem 6

Question

Find the sum of all positive integers

n

such that

\sqrt{n^2 + 85n + 2017}

is an integer.

Answer

Set

n^2 + 85n + 2017 = m^2

and multiply by

4

so the left side becomes

(2n + 85)^2 + 843

Factor

4m^2 - (2n + 85)^2

as a difference of squares;

843 = 3 \cdot 281

has only two factorizations

Suppose

n^2 + 85n + 2017 = m^2

for a positive integer

m.

Multiplying by

4

and completing the square gives

(2n + 85)^2 + 843 = 4m^2,

so

\begin{aligned} &(2m - 2n - 85)(2m + 2n + 85) \\ &= 843 = 3 \cdot 281, \end{aligned}

where

281

is prime. Both factors are positive with the second one larger, so either

2m - 2n - 85 = 1

and

2m + 2n + 85 = 843,

or

2m - 2n - 85 = 3

and

2m + 2n + 85 = 281.

The first system gives

m = 211

and

n = 168,

and indeed

168^2 + 85 \cdot 168 + 2017

= 44521 = 211^2.

The second gives

m = 71

and

n = 27,

with

27^2 + 85 \cdot 27 + 2017

= 5041 = 71^2.

The requested sum is

168 + 27 = 195.

Problem 6 in other years