for all positive integers m there exists consecutive primes which are at least m apart I'm having difficulty as to how I should approach this problem, any help would me much appreciated!
Note that $k$ divides $n! + k$ for each $k\le n$. Use this fact to show that for all positive integers $m$ there exist consecutive primes which are at least $m$ apart. 
 A: $$(m+1)!+2, ......, (m+1)!+(m-1), (m+1)!+m, (m+1)!+(m+1)$$ are all non-primes.
A: It's trivially true that $1|n! + 1$. Far more useful is the fact that $2|n! + 2$ for all $n > 1$. This means that $n! + 2$ must be even, and therefore composite. Likewise $n! + 3$ is divisible by $3$ for all $n > 2$. Where I'm going with this is that $k$ is a nontrivial divisor of $n! + k$ for each $1 < k \leq n$. If we want at least $m$ consecutive composite numbers, we set $n = m + 1$. (In some cases this might give us a few more than we want).
For example, if we want ten consecutive composite numbers, we set $n = 11$ and verify that:


*

*$11! + 1$ is prime

*$11! + 2$ is even

*$11! + 3$ is divisible by $3$

*$11! + 4$ is divisible by $4$

*$11! + 5$ is divisible by $5$

*$11! + 6$ is divisible by $6$

*$11! + 7$ is divisible by $7$

*$11! + 8$ is divisible by $8$

*$11! + 9$ is divisible by $9$

*$11! + 10$ is divisible by $10$

*$11! + 11$ is divisible by $11$

*$11! + 12$ is divisible by $3$ and $4$

*$11! + 13$ is... you can check this one yourself.

