how can we find the number of positive integers n for which n^2+96 is a perfect square?
Asked by | 26th 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 - n m + 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.
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 - n m + 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 | 26th Aug, 2012, 06:50: PM
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