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Find 3 consecutive positive odd integers such that the sum of the squares of the first two is 15 less than the square of the third?
MikeOxoar
n^2 + (n+2)^2 = (n+4)^2 - 15
2n^2 + 4n +4 = n^2 + 8n +16 - 15
n^2 - 4n + 3 = 0
(n - 1)(n - 3) = 0
n = 1 or n = 3
1 and 3 both fulfill the condition, because
1^2 + 3^2 = 5^2 - 15
i.e. 1 + 9 = 25 - 15 = 10
and
3^2 + 5^2 = 7^2 - 15
i.e. 9 + 25 = 49 - 15 = 34
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