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how can we find the number of positive integers n for which n^2+96 is a perfect square?
Asked by | 26 Aug, 2012, 06:39: PM Expert Answer

Suppose that n2 + 96 = m2 for some positive integer m, then

m2 - n2 = 96
(m - n)(m + n) = 96 = 25 × 3

96 has 12 divisors which come in pairs and since m - n < m + n they are

m - nm + n
1 96
2 48
3 32
4 24
6 16
8 12

For each of the 6 cases solve for m and n and verify which pairs, if any, satisfy your requirements.

Answered by | 26 Aug, 2012, 06:50: PM

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