elementary number theory - 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

Answer

The sum of the integers from $1$ to $n$ is $n(n+1)/2$. The sum of the integers from $k$ to $k+n$ is then
$$\begin{align*}
k+(k+1)+\cdots+(k+n) &= (n+1)k + 1+\cdots+n\\
& = (n+1)k + \frac{n(n+1)}{2} \\
&= \frac{(n+1)(2k+n)}{2}.\end{align*}$$

Therefore, $a$ can be expressed as the sum of consecutive integers if and only if $2a$ can be factored as $(n+1)(2k+n)$.

Suppose that $a$ is a power of $2$. Then $2a$ is a power of $2$, so $(n+1)(2k+n)$ must be a power of $2$. If we want to avoid negatives, and also avoid the trivial expression as a sum with one summand, we must have $n\geq 1$ and $k\gt 0$. But the parities of $n+1$ and of $2k+n$ are opposite, so this product cannot be a power of $2$ unless either $n+1=1$ (which requies $n=0$) or $2k+n=1$ (which requires $k=0$). Thus, no power of $2$ can be expressed as a sum of at least two consecutive positive integers. In particular, $8$, $16$, $32$, etc cannot be so expressed.

On the other hand, suppose that $a$ is even but not a power of $2$. If we can write $2a = pq$ with $p\gt 1$ and odd, $q$ even, and $q\geq p+1$, then setting $n=p-1$ and $k=(q-p+1)/2$ gives the desired decomposition. If this cannot be done, then every time we factor $2a$ as $pq$ with $p\gt 1$ odd, we have $q\lt p+1$. Then we can set $n=q-1$ and $k = (p+1-q)/2$.

Thus, the powers of $2$ are the only even numbers that are not expressible as the sum of at least two consecutive positive integers.

Added. The OP has now excluded powers of $2$, but has also required that the sum contains strictly more than two summands; i.e., $k\gt 0$ and $n\gt 1$. With the above decompositions, the only case in which we could have $n=1$ is if $2a=pq$ with $p$ odd, $p\gt 1$, and $q=2$. But this is impossible, since $a$ is assumed to be even, and this leads to $2a = 2p$ with $p$ odd.