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HMMT February

Number theory Difficulty 4.9 AIME Find the answer

Compute the sum of all positive integers nn such that n2−3000n^{2}-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=x2n^{2}-3000=x^{2}, so n2−x2=3000n^{2}-x^{2}=3000. This factors as (n−x)(n+x)=3000(n-x)(n+x)=3000. Thus, we have n−x=2an-x=2a and n+x=2bn+x=2b for some positive integers a,ba, b such that ab=750ab=750 and a<ba<b. Therefore, we have n=a+bn=a+b, so the sum will be just the sum of divisors of 750=2⋅3⋅53750=2 \cdot 3 \cdot 5^{3}, which is (1+2)(1+3)(1+5+25+125)=1872(1+2)(1+3)(1+5+25+125)=1872.

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