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I'm trying to prove that every non-prime natural number greater than $1$ can is equal to a sum of consecutive even or odd numbers.
This can be resumed in : « $p,m,n \in ℕ$» , «$p > 1$» , «$n > 0$» and «$ m>0$».

The problem is :

Prove that « $p = n+(n+2)+(n+4)+...+(n+2m)$ ».

So far, I only managed to prove it for number such as $p = 2a+2$ with $a \in ℕ$.

Can you please help me ?

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    $\begingroup$ So, $$p=(m+1)(m+n)$$ $\endgroup$ – lab bhattacharjee Jul 29 '14 at 12:41
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    $\begingroup$ I would suggest a different letter from $p$ when you wish to speak of "a natural number, which, by hypothesis, is not prime." $\endgroup$ – hunter Jul 29 '14 at 12:42
  • $\begingroup$ If $p$ is composite $p = ab$ ($1<a\leq b$). Take $m=a-1$ and $n = b-m$ and use lab bhattacharjee's comment above. $\endgroup$ – Winther Jul 29 '14 at 12:48
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Let $p = ab$.

wlog let $a \le b$.

Set $m = a - 1$, and $n = b - m = b + 1 - a$.

Now the sum of your series = $(n+m)(m+1) = ab = p.$

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If $p$ is even, then the solution is obvious: $p=2k = (k-1) + (k+1)$.

You can pull a similar trick if $p$ is divisible by $3$: $p=3k = k-2+k+k+2$

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    $\begingroup$ It's rather $3k= (k-2)+k+(k+2)$. Now you can generalize your argument for a factorization $p=kl$ with $k$ and $l$ odd. $\endgroup$ – Quang Hoang Jul 29 '14 at 12:43

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