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.

Topic

Number Theory

Difficulty Level

8.5 out of 10

Suggested Book

Elementary Number Theory by David M. Burton

Start with hints

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 \(l\ge 0\). (Proved).

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