I need to prove this using the Chinese Residue Theroem. If i choose n even (n>2) then all even number are composite of course, but for the odd i don't know what to do. Plus, i don't see how the chinese residue theroem could be helpful. Thanks for your hints.
I know how to prove that there is infinitely prime (and infinitely composite) and i think i will have to use this.
On
Take $n=k!+2$ Then all numbers $k!+2,k!+3,\ldots ,k!+m$ with $1\le m\le k$ are composite. From this statement your claim follows.
Reference: Prove that $n! + k$ is a composite number
Hint : Try to prove that for all $k\geq 0$, $$n=(1002+k)!+2$$
satisfy the property.