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.
Best Answer
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.