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The sum of more than two consecutive natural numbers cannot be prime.

Is the statement true and is there any way to prove it?

I was able to prove that the sum of an odd amount of consecutive numbers cannot be prime:

So, since the sum of consecutive integers is $x+(x+1)+(x+2)+(x+3)$ etc... we can also write this as $$nx + n(n-1)/2 = n(x + (n-1)/2)$$ with $n$ as the amount of numbers and $x$ the first number in the row. So, with an odd number as $n\neq 1$, we will get a product which will never result in a prime.

Any way to prove this for all $n \ge 2$? Thanks for all the help.

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  • $\begingroup$ x instead of n if the 1st expression and after, the bounds for xn ?? $\endgroup$ – user354674 Jul 18 '16 at 17:58
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    $\begingroup$ Sometimes when you prove something for odd numbers you just need to prove it for even numbers and then you're done. $\endgroup$ – Robert Soupe Jul 19 '16 at 3:31
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For a sum of three or more consecutive positive integers

$S = x + (x+1) + (x + 2) + ..... + (x + n -1)$ $x > 0; n > 2$

$S = (x + n-1) + (x+n - 1) + (x + n - 2) + ..... + (x + 1)+x$

Add 'em together.

$2S = (2x + n -1) + (2x + n-1) + .... (2x + n-1) = n(2x + n-1)$

Case one: $n$ is even. Then $S=\frac n2(2x + n -1)$ is not prime as $n/2 > 1$ and $2x + n - 1 > 2$

Case 2: $n$ is odd. Then $2x + n - 1$ is even and $S = n\frac{2x + n - 1}2$ which is not prime as $n > 1$ and $(2x + n - 1)/2 > 1$.

You were 90% of the way there. You just needed to hit it with your paddle a few more times.

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Let our sum be $$p=nx+n (n+1)/2\qquad n>2$$ If $n $ is odd then $(n+1)/2$ is an integer and we can write $p=nk $, $k\in\mathbb {N} $ making $p $ composite. If $n $ is even then $n/2=k'$ and we can write $p $ as $$p=k'(2x+[n+1])=k'd$$ $d\in\mathbb {N} $ again making $p $ composite. Note that we must have $n>2$ because otherwise some of our factors will be one which don't count as factors in making a number composite.

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The sum is as you said $nx + n(n-1)/2$. If $n$ is odd it is a multiple of $n$ as you said.

Yet, if $n$ is even it is a multiple of $n/2$.

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