I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.

Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.


Hence, $g=kx+\frac{k(k-1)}{2}$

Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$

So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:

Case 1)

$k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.

Case 2)

$k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.

Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.

If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.

If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.

But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.

Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.

Any feedback on this proof would be appreciated.

  • 1
    $\begingroup$ I had to look around to find the justification for "if $k$ is even then $k$ cannot divide $g$" which is that $x=g/k +(k-1)/2 $. (If $k|g$ and $k$ is even then $(k-1)$ and $x$ are not integers.) There is a flaw: The result is (trivially) false if you don't require $k>1$: In the case where $k$ is odd ,$ k$ divides$ j$ but $j$ DOES have an odd factor: the number $1$. So if $k$ is odd then $k=1.$..... But other than this, it's a valid proof. $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:01
  • $\begingroup$ An alternate proof: Assume $k>1. $ A sum of $k$ consecutive positive integers is $a(a+1)/2- b(b+1)/2$ where $a,b$ are positive integers with $ a-b=k\geq 2.$ Now $$a(a+1)/1-b(b+1)/2= ((a+1/2)^2-1/4)/2-((b+1/2)^2-1/4)/2=((a+1/2)^2-(b+1/2)^2)/2=(a+b+1)(a-b)/2.$$ If this equals $2^n$ then $(a+b+1)(a-b)=2^{n+1}.$ But one member of $S=\{a+b+1,a-b\}$ is odd and the other member is even, and they're both positive and their product is a power of two. This requires that the smaller member of $S$ is equal to $1$. That is, $a-b=1,$ contrary to $a-b\geq 2.$ $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:14
  • $\begingroup$ Are we allowed $k=1$? :-) $\endgroup$ – Joffan Feb 27 '17 at 23:19
  • $\begingroup$ @Joffan. In day-to-day discourse, a consecutive number of things is usually only said when there are at least 2 of them, otherwise we usually say "a thing". (Except for the Toronto Maple Leafs, who occasionally have a winning streak of 1 consecutive game.).... In mathematics, it depends on the context, or the preference of the writer, so it's best to be explicit at the beginning. $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:25

It is a bit slicker to do it in the other direction: Instead of starting with $g$ and looking for a way to write it as the sum of consecutive integers, start by a sum of $k\ge 2$ consecutive integers and show that its sum cannot be a power of two.

We use the general fact that the sum of a finite arithmetic sequence is $$ (\text{number of terms})\frac{(\text{first term})+(\text{last term})}{2} $$

If there's an odd number of terms ($\ge 3$), then the first factor of this is divisible by some odd prime, and dividing by $2$ at the end cannot make that go away. So the sum is divisible by the same odd prime, and is not a power of $2$.

If there's an even number of terms, then exactly one of the first and last terms will be odd. So the numerator in the fraction above is odd, and is divisible by an odd prime; like before this odd prime will divide the entire expression.

  • $\begingroup$ Good, except, as I said in a comment, there is the unspoken assumption that $k>1.$..............+1 $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:19
  • $\begingroup$ @user254665: Are you saying I should have left it unspoken? Why? $\endgroup$ – Henning Makholm Feb 27 '17 at 23:32
  • $\begingroup$ I should have said the assumption In the Q, not in your A. $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:40

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