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Show that any integer that is not a power of 2 can be written as the
sum of consecutive integers.
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Well, as stated this is trivial to prove. For any positive integer n add the numbers -(n - 1) + -(n - 2) + ... + -1 + 0 + 1 + ... + (n - 2) + (n - 1) + n. Then every positive number less than n is cancelled by it's negative counterpart and the sum of the series is n. This works for any positive n, even the powers of two, and can be easily reversed for negative values of n.
Assuming that you meant a sum of consecutive positive integers, consider the series n - k, n - (k - 1), n - (k - 2), ..., n - 1, n, n + 1, ..., n + (k - 1), n + k. Adding these together cancels out all of the values except the n's, of which there are 2k + 1. The sum of those numbers will be n(2k + 1) = S, where n and k can be any positive integers. If S is a composite number that is not a power of 2, then let 2k + 1 be one of it's odd prime factors and n represent the rest of its prime factors. If S is prime then let n = 1 and 2k + 1 = S.
In cases where 2k + 1 > 2n the series will stretch into the negatives, but since those negatives will cancel with their positive counterparts we can still construct a series of positive integers only that will add up to S. For example, for S = 5 we would have n = 1 and k = 2, and our series would be -1 + 0 + 1 + 2 + 3. However, the -1 and 1 cancel out, and the 0 can be ignored, so our series will be 2 + 3. This also shows us why we can't find any such series for a power of 2: 2k + 1 is always odd, and powers of 2 have no odd prime factors.
Wrap it in bacon
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