Integers as consecutive sums

Starting to check with some examples, we notice quickly that it seems like only the powers of two cannot be expressed this way.

• To prove this, suppose first that our integer $a=2^k$ can be written as a sum of consecutive integers :

$$2^k=n+(n+1)+\ldots+(n+m)=mn + \frac{(m-1)m}{2}$$

with $m\geq 2$. Then by multiplying by 2 to get rid of the denominator :

$$2^{k+1}=2mn + (m-1)m$$

which shows that $m$ divises $2^{k+1}$, hence $m$ is a power of 2 itself, say, $m=2^r$, with $r\geq 1$.
Plugging this back into the equation gives :

$$2^{k+1}=2^{r+1} n+\left(2^r-1\right) 2^r=2^r\left(2 n+2^r-1\right)$$

But this is impossible : the term $2n+2^r-1$ is odd, so it cannot be a factor of $2^{k+1}$.
This contradictions means that $2^k$ can never be written as a sum of consecutive integers.

• Now suppose that our integer $a$ is not a power of 2. This means that $a$ has an odd factor, say $q$, such that $a=2^k q$ for some $k\geq 0$. Let’s show that it can be written as a sum of consecutive integers. We work backwards from the answer : assume that it can be written this way, that is, there is a positive integer $n$ and $m\geq 2$ such that :

$$a=2^{k}q=n+\ldots+(n+m-1)=mn+\frac{m(m-1)}{2}$$

And again multiplying by 2 on each side to get rid of the denominator :

$$2^{k+1} q=m(2 n+m-1)$$$m$ and $2n+m-1$ are of opposite parity. Now, a natural choice is to suppose one of the two following :

1. $2^{k+1}=m$ and $2n+m-1=q$. We get that $n=\frac{q+1}{2}-2^{k}$. This is positive for $\frac{q+1}{2}\geq 2^{k}+1$. If this condition is verified, let’s check that this choice of $m$ and $n$ yields the right decomposition :

$$n+\ldots+(n+m-1)=m n+\frac{m(m-1)}{2}=2^{k+1}\left(\frac{q+1}{2}-2^k\right)+\frac{2^{k+1}\left(2^{k+1}-1\right)}{2}=2^k q=a$$2) If $2^{k+1}=2n+m-1$ and $q=m$, we get that $n=2^{k}-\frac{q-1}{2}$. This is positive for $\frac{q+1}{2} \leqslant 2^k$. Again, it is straightforward to check that if this condition is verified, then this choice of $n$ and $m$ are correct.

Because one of the two conditions has to be true, then the property has been shown.