 ## Understand the problem

Prove that the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r>1$ ,are of the form $n=2^l$ for some $l\ge 0$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.

Number Theory

8.5 out of 10

##### Suggested Book

Do you really need a hint? Try it first!

Claim: Any positive integer $n$ can be written as $n=2^k\cdot m$ , where $k\ge0$ and $m$ is an odd positive integer. To prove this claim use the fact :  $n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}$, where $k$ and all $k_i$ are non-negetive integers and all $p_i$ are odd primes.

The sum of (any) $r$ consecutive positive integers is given by, $(q+1)+(q+2)+(q+3)+\cdots+(q+r)$ =   $q\cdot r+(1+2+3+\cdots+r)$ =   $q\cdot r+\frac{r(r+1)}{2}$.

Equating this sum to $n$ we get,           $2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}$   Or,     $2^{k+1}\cdot m = 2q\cdot r+r(r+1)$   Or,     $2^l\cdot m = r(2q+r+1)$ , where $l=k+1\ge1$.

In particular if we take $n=2^l$ then $m$ is equal to 1. Since both $r$ and $(2q+r+1)$ are greater than 1, so they can’t be equal to $m$ in this case. Again one of $r, (2q+r+1)$ is odd integer which implies the product $r(2q+r+1)$ can’t be equal to $2^l$.  $\Rightarrow 2^l \neq r(2q+r+1)$ $\Rightarrow 2^l$ can’t be expressed as the sum of $r$ consecutive positive integers with $r>1$ and $l\ge 1$. Now, $n=2^0=1$ also can’t be written in the same manner when $l=0$. Therefore the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r>1$ , are of the form $n=2^l$ for some $l\ge 0$. To show that any number other than of the form $2^l$ is sum of consecutive integers:
Observe that for $n=2^l.m$, where $m$ is an odd number greater than 1, there are two cases:
1. $m < 2^l$
Select $r$ and $q$ such that $r = m$ and $2q+r+1 = 2^l$. 2. $m > 2^l$
Select $r$ and $q$ such that $2q+r+1 = m$ and $r = 2^l$. (QED).

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