Compute the sum of all positive integers n such that n2−3000 is a perfect square.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Suppose n2−3000=x2, so n2−x2=3000. This factors as (n−x)(n+x)=3000. Thus, we have n−x=2a and n+x=2b for some positive integers a,b such that ab=750 and a<b. Therefore, we have n=a+b, so the sum will be just the sum of divisors of 750=2⋅3⋅53, which is (1+2)(1+3)(1+5+25+125)=1872.
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