2017 AIME II Problem 6

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Concepts:Diophantine Equationcompleting the squaredifference of squares

Difficulty rating: 2450

6.

Find the sum of all positive integers nn such that n2+85n+2017\sqrt{n^2 + 85n + 2017} is an integer.

Solution:

Suppose n2+85n+2017=m2n^2 + 85n + 2017 = m^2 for a positive integer m.m. Multiplying by 44 and completing the square gives (2n+85)2+843=4m2,(2n + 85)^2 + 843 = 4m^2, so (2m2n85)(2m+2n+85)=843=3281,(2m - 2n - 85)(2m + 2n + 85) = 843 = 3 \cdot 281, where 281281 is prime. Both factors are positive with the second one larger, so either 2m2n85=12m - 2n - 85 = 1 and 2m+2n+85=843,2m + 2n + 85 = 843, or 2m2n85=32m - 2n - 85 = 3 and 2m+2n+85=281.2m + 2n + 85 = 281.

The first system gives m=211m = 211 and n=168,n = 168, and indeed 1682+85168+2017=44521=2112.168^2 + 85 \cdot 168 + 2017 = 44521 = 211^2. The second gives m=71m = 71 and n=27,n = 27, with 272+8527+2017=5041=712.27^2 + 85 \cdot 27 + 2017 = 5041 = 71^2.

The requested sum is 168+27=195.168 + 27 = 195.

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