Prove that all even integers $n \neq 2^k$ are expressible as a sum of consecutive positive integers How do I prove that any even integer $n \neq 2^k$ is expressible as a sum of positive consecutive integers (more than 2 positive consecutive integer)?
For example: 
14 = 2 + 3 + 4 + 5
84 = 9 + 10 + ... + 15

n = sum (k + k+1 + k+2 + ...) 
n ≠ 2^k


(The following is merely a simplification of @Arturo Magidin's proof. So, please do not upvote my post.) Suppose $S=k+(k+1)+\ldots+\left(k+(n-1)\right)$ for some $k\ge1$ and $n\ge2$. Then
$$
S = nk + \sum_{j=0}^{n-1} j = nk+\frac{n(n-1)}{2} = \frac{n(n+2k-1)}{2}.
$$
Hence $S\in\mathbb{N}$ can be written as a consecutive sum of integers if and only if $2S = nN$ with $N\ (=n+2k-1) >n\ge2$ and $N, n$ are of different parities.
If $S$ is a power of $2$, so is $2S$. Hence the above factorization cannot be done.
If $S$ is not a power of $2$ (whether it is even is immaterial), we may write $2S=ab$, where $a\ge3$ is odd and $b\ge2$ is a power of $2$. Therefore we can put $N=\max(a,b)$ and $n=\min(a,b)$.
