# Proving step to full proof

• Oct 1st 2010, 07:57 PM
I-Think
Proving step to full proof
I was doing a proof to prove that there exists arbitrarily long strings of composite numbers.
In more technical language, for every $\displaystyle m\in {N}$, there exists a n such that
$\displaystyle n, n+1, n+2, n+3,..., n+m$ is composite

to do so, I let $\displaystyle n=m!$,(for m>3) then we get the sequence:
$\displaystyle m!, m!+1, m!+2, m!+3,..., m!+m$

Every element of this sequence is obviously divisible by some integer, except $\displaystyle m!+1.$

Now I must prove that m!+1 is always an composite. I have no idea how to do this. I tried induction and contradiction.
• Oct 1st 2010, 10:53 PM
CaptainBlack
Quote:

Originally Posted by I-Think
I was doing a proof to prove that there exists arbitrarily long strings of composite numbers.
In more technical language, for every $\displaystyle m\element{N}$, there exists a n such that
$\displaystyle n, n+1, n+2, n+3,..., n+m$ is prime

to do so, I let $\displaystyle n=m!$,(for m>3) then we get the sequence:
$\displaystyle m!, m!+1, m!+2, m!+3,..., m!+m$

Every element of this sequence is obviously divisible by some integer, except $\displaystyle m!+1.$

Now I must prove that m!+1 is always an integer. I have no idea how to do this. I tried induction and contradiction.

There are so many typos in this I think you should correct them before expecting a reply.

Also this is number theory.

To show that for any $\displaystyle n\in \mathbb{N}$ there is a set of $\displaystyle n$ consecutive composites choose $\displaystyle $$m so that \displaystyle m=n+1. Then \displaystyle m!+2, m!+3,.. m!+m are all composite, and there are \displaystyle m-1=n of these. CB • Oct 1st 2010, 10:59 PM CaptainBlack Quote: Originally Posted by I-Think Now I must prove that m!+1 is always an composite. I have no idea how to do this. I tried induction and contradiction. Help please You can't since if \displaystyle$$ m=2$ then $\displaystyle m!+1=5$ is prime

CB
• Oct 2nd 2010, 07:50 AM
I-Think
I apologize for the typos. I also apologize for posting this in the wrong forum but I did not expect the problem to involve Number Theory, as it is a topic I had not yet covered.

And as for proving that $\displaystyle m!+1$is always composite, I had stated it was over the range $\displaystyle m>3$
• Oct 2nd 2010, 09:40 AM
Traveller
Let n = (m+2)! + 2

P.S: m! + 1 is composite when m = p-1 for some prime p > 3.
• Oct 2nd 2010, 02:35 PM
CaptainBlack
Quote:

Originally Posted by i-think
i apologize for the typos. I also apologize for posting this in the wrong forum but i did not expect the problem to involve number theory, as it is a topic i had not yet covered.

And as for proving that $\displaystyle m!+1$is always composite, i had stated it was over the range $\displaystyle m>3$

ok 11!+1

cb
• Oct 3rd 2010, 11:12 PM
aman_cc
Infact following should work

n = (m+1)! + 2