Summer Problem Solving Marathon Solution #6

Since n + 125 and n + 199 are both perfect squares, we have n + 125 = a² and n + 199 = b² for some a and b. Subtracting the first equation from the second, we have:

b² - a² = 76

or:

(b + a)(b - a) = 76

This yields three possibilities:

#1: b + a = 76, b - a = 1
#2: b + a = 38, b - a = 2
#3: b + a = 19, b - a = 4

But only the second yields an integer value for b. So b = 20, and a = 18. So n + 201 = 20² = 400, and n = 199.