Is it true that odd primes can be expressed only one way as a sum of Consecutive Counting Numbers?

How do i prove this to be true?Is it true that odd primes can be expressed only one way as a sum of Consecutive Counting Numbers?
Well, there is the trivial problem:





17 = 17 (one consecutive)


17 = 9 + 8.





So under the assumption that the sum is nontrivial:





The sum (n %26gt; m + 1 %26gt; m 鈮?0)





(m+1) + (m+2) ... + (n-1) + n


= (1/2) (n(n+1) - m(m+1))





follows from





1 + 2 + .. + n = n(n+1) / 2.





Let p be an odd prime. Then





2p = n(n+1) - m(m+1)


= n虏 + n - m虏 - m


= (n虏 - m虏) + (n - m)


= (n - m) (n + m + 1)





p cannot be two since there is an odd factor on the RHS. If p is an odd prime, one of





n - m and n + m + 1





must be 2, with the other equalling p.





If n + m + 1 = 2 鈫?n + m = 1 鈫?not possible since n %26gt; m + 1 %26gt; 0.





Therefore





n - m = 2


n + m + 1 = p





That is,





p = floor(p/2) + ceiling(p/2)





is the only nontrivial sum of consecutive numbers adding to the odd prime p.





Hope this helps.





鈾?鈾?br>




Edit: A little refinement.Is it true that odd primes can be expressed only one way as a sum of Consecutive Counting Numbers?
5 is an odd prime, I can write it as 1 + 2 + 2, so this is not true.