Number of ways to write n as a sum of consecutive integers

Question

Allow $d(n)$ to be the number of divisors of $n$. Show that there are $d(m)$ ways to write $n$ as the sum of consecutive integers where $m$ is the largest odd divisor on $n$. I have absolutely no idea where to go. The hint given to me is that $$\ x + (x+1) +... + (x+r-1) + (x+r) = (r+1)x + \frac{1}{2}r(r+1)$$ I can set the above expression equal to $n$, but this still confuses me.

Answer 1

Write $n=2^q m$ where $m$ is the largest odd integer which divides $n$ and $q\geq 0$. We will use the formula given in the hint in the following form

$$\sum_{i=0}^{2r} x+i=(2r+1)(x+r).$$

Suppose we have a factorization of $m$ as $d_1 d_2$, both them are odd. Take

$$x=2^q d_2-\frac{d_1-1}{2}=2^q \frac{m}{d_1}-\frac{d_1-1}{2},$$

and $r=(d_1-1)/2$. A simple computation shows that $(2r+1)(x+r)=n$. Hence, for each divisor of $m$ we found a sequence of numbers which sum is $n$ and all are diferents.

Answer 2

As written, this is demonstrably false.

Consider $n=3$, $m=3$, $d(m)=2$ but

$$3=3$$ $$1+2=3$$ $$0+1+2=3$$

3 ways, not 2.

Similarly $n=5$, $m=5$, $d(m)=2$ but

$$5=5$$ $$2+3=5$$ $$-1+0+1+2+3=5$$

3 ways, not 2.