Prove that (2n+3)2−(2n−3)2(2n + 3)^2 - (2n - 3)^2(2n+3)2−(2n−3)2 is always a multiple of 12, for all positive integer values of nnn.
Hence prove that the difference between the squares of any two odd positive integers that differ by 6 is always a multiple of 12.
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