Let n be a postive integer greater than 1. Show that n can be written as a sum of at least 2 consecutive postive integers if and only if n is not a power of 2.

First, we'll show that if n is not a power of 2, then it can be written as a sum of at least 2 consecutive positive integers.

If n is an odd number, say n = 2k + 1, then n = k + (k+1), and since n is at least 3, k is at least 1, and we have written n as a sum of 2 consecutive positive integers.

If n is an even number, but not a power of 2, then it has an odd factor,
so we can write n as 2^{k}s, where s is an odd number greater
than or equal to 3.
Let s = 2r + 1. Then:

n = (2^{k} - r) + (2^{k} - r + 1) + ... + (2^{k} + r - 1) + (2^{k} + r)

In the decomposition above, there may be negative or zero terms. But note that:

(2^{k} + r) - (r - 2^{k}) = 2^{k+1} > 1

So if there are negative terms, then for each one there is a
corresponding positive terms, plus there are at least 2^{k+1}
postive terms left over.
Hence we can just remove any zero term, and pairs of matching
positive and negative terms, and we're left with n as the sum of
at least two consecutive positive integers.

To finish the proof, we must show that if n can be written as a sum of consecutive integers, then it is not a power of 2. Let:

n = a + (a+1) + (a+2) + ... + (a+t)

where t > 0. This can be rewritten as:

n = (t+1)a + t(t+1)/2 = (t+1)(2a + t)/2

If t is even, then n is a multiple of t+1, which is odd, so n cannot be a power of 2. If t is odd, then 2 divides evenly into t+1, so n is a multiple of 2a+t, which is odd, so again n cannot be a power of 2.