Prime Sums

Number Theory Level 5

If $p$ is an odd prime number, at most how many ways can $p$ be expressed as a sum of two or more consecutive integers?

\lfloor \sqrt{p} \rfloor

3 1 4 2

p - \phi(p)

None of the choices

2 solutions

Efren Medallo
May 12, 2017

For all odd primes $p$ , there are only $3$ ways for it to be expressed as a sum of $k$ consecutive integers.

$k = 2$ , since $p$ is odd, then it will be easy to prove this out.
$k = 2p$ , and that specifically is the sequence of integers from $-(p-1)$ to $p$ .
$k = p$ , and that is the sequence of integers from $-(\frac{p-3}{2})$ to $\frac{p+1}{2}$ .

I will post the proof a little while later.

So please tell me how can be 5 expressed as sum of two or more consecutive integers in 3 ways?

akash patalwanshi - 4 years ago

$5 = 2+3$

$5 = -4 + -3 + -2 + -1 + 0 + 1 +2 + 3 + 4 + 5$

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

Efren Medallo - 4 years ago

Thank u. I didn't noticed question was asked for whole integers. I consider only positive integers

akash patalwanshi - 4 years ago

Jason Martin
Nov 13, 2017

We have that $p=n+(n+1)+(n+2)+\ldots +(n+m-1)=mn+\frac{m(m-1)}{2} \Rightarrow 2p=m(2n+m-1)$ . In particular, $m|2p \Rightarrow m=2, p,$ or $2p$ (because there must be $2$ or more integers, $m \geq 2$ ). This yields exactly $\boxed{3}$ solutions for $(m,n)$ .

Note from this, we find $n=\frac{p-1}{2}, \frac{p+1}{2},$ or $-(p-1)$ respectively. As an example of $n=-(p-1)$ , consider $p=5$ . Then $-4-3-2-1+0+1+2+3+4+5=5$

Prime Sums

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