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I proved that integers of the form 2^a can't be expressed as a sum of r consecutive integers but I have no idea how to prove the above statement

Raunit Singh
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    A number can be expressed as $r$ consecutive integers if it is of th form $$\frac{r(2n+r-1)}2$$ for $n$ the starting integer. – Thomas Andrews Sep 07 '19 at 23:25
  • Note, this means that $r$ must be the odd factor. If you have $3$ as a factor of $m$ you might not be able to write $m$ as the sum of $r=5$ consecutive integers. – Thomas Andrews Sep 07 '19 at 23:28
  • @ThomasAndrews: How does $5=2+3$, being the sum of $r=2$ consecutive integers, fit what you have said? I know $5=-1+0+1+2+3$ also works, but still – Henry Sep 07 '19 at 23:34
  • @henry Not sure what your question is. It’s not an “if and only if” statement, only that if $m$ is divisible by odd $r$ then $m$ can be written as the sum of $r$ consecutive integers. $r=2$ is irrelevant because $2$ isn’t odd. – Thomas Andrews Sep 07 '19 at 23:37
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    @ThomasAndrews I interpret the formula as saying $r$ can be even, but then $2n+r-1$ is an odd factor of the sum. – David K Sep 08 '19 at 00:03
  • @DavidK The question doesn’t really say what $r$ is, but if $m$ is divisible by an odd integer, it is not necessarily true than $m$ can be written as the sum of $r=2$ consecutive integers. For example, $m=6.$ It is true, though, if $m$ is divisible by an odd integer $r,$ then $m$ can be written as the sum of $r$ consecutive integers. – Thomas Andrews Sep 08 '19 at 00:13
  • @ThomasAndrews If $m=6$ then $r$ cannot be $2,$ but $r=4$ is possible: $0+1+2+3=6.$ What I get from the formula is that in general if the sum is even then $r \not\equiv 2\pmod 4,$ but $r=4$ is possible and $r$ may be a larger multiple of $4$ as long as $r \leq 2m.$ For example, $(-5)+\ldots+6=6$ with $r=12.$ Of course $r$ can always be an odd factor of $m$ instead. Saying that one thing is possible is not saying another thing is impossible. – David K Sep 08 '19 at 00:45
  • See also https://math.stackexchange.com/a/139852/589 – lhf Sep 08 '19 at 01:54

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