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

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.

• x instead of n if the 1st expression and after, the bounds for xn ?? – user354674 Jul 18 '16 at 17:58
• Sometimes when you prove something for odd numbers you just need to prove it for even numbers and then you're done. – Robert Soupe Jul 19 '16 at 3:31

For a sum of three or more consecutive positive integers

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

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

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

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.

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.

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 $n/2$.