Positive Integer is Sum of Consecutive Positive Integers iff not Power of 2

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then $n$ can be expressed as the sum of $2$ or more consecutive (strictly) positive integers $n$ is not a power of $2$.

Necessary Condition
Let $a$ be the smallest of $m$ consecutive (strictly) positive integers, where $m \ge 2$.

From Sum of Arithmetic Sequence, their sum is $\dfrac {m \paren {2 a + m - 1} } 2$.

$\dfrac {m \paren {2 a + m - 1} } 2$ is a power of $2$.

Then $m$ and $2 a + m - 1$ must also be powers of $2$.

Since $m \ge 2$, $m$ must be even.

Then $2 a + m - 1$ is odd.

Together with $2 a + m - 1$ being a power of $2$, we have $2 a + m - 1 = 1$.

However, $2 a + m - 1 > 2 \times 0 + 2 - 1 = 1$.

This is a contradiction.

Therefore any sum of $2$ or more consecutive integers cannot be a power of $2$.

Sufficient Condition
Let $n$ be an integer that is not a power of $2$.

Then $n$ contains an odd factor greater than $1$.

Let $d = 2 m + 1$ be such a factor.

Then:
 * $\dfrac n d - m, \dfrac n d - m + 1, \dots, \dfrac n d + m$

is a sequence of $2 m + 1$ consecutive integers.

From Sum of Arithmetic Sequence:

It may happen that $\dfrac n d - m \le 0$.

In that case, notice that:


 * $\dfrac n d - m \le m - \dfrac n d \le \dfrac n d + m$

and so $m - \dfrac n d$ is also (somewhere) in our sequence of $2 m + 1$ consecutive integers.

So all negative integers in our sequence of $2 m + 1$ consecutive integers cancel out with their positive counterparts.

Hence we can remove all numbers from $\dfrac n d - m$ to $m - \dfrac n d$.

As a result of that process, the sum remains unchanged.

We see that the number of integers eliminated is $2 \paren {m - \dfrac n d} + 1$.

Since $\dfrac n d$ is a factor of $n$, we have $\dfrac n d \ge 1$.

Hence the number of integers remaining is:

This finishes our proof.