# How to prove that there are infinitely many integers n for which each of n, n+1, . . . ,n+ 1000 is composite. [duplicate]

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.

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.

• Thanks for your anwser, but i need to use the Chinese Residue Theorem... Also this would be a good proof, but i also need n+1 and this one starts at n+2
– R-B
Oct 20 '20 at 18:57
• @R-B No, it starts with $n,n+1,n+2,\ldots,$, right? We have $k!+3=(k!+2)+1=n+1$. Oct 20 '20 at 21:59
• But yes, you are right. There is also a nice solution using CRT, see the duplicate Bill pointed out.. Oct 21 '20 at 8:16

Hint : Try to prove that for all $$k\geq 0$$, $$n=(1002+k)!+2$$

satisfy the property.

• Where does that n comes from?
– R-B
Oct 20 '20 at 18:46
• From my mind. The question is to find $n$ such that, so you can drop it and verify that it satisfies the given property. Oct 20 '20 at 18:47
• @TheSilverDoe Yes, but from which portion of your mind, the left parietal lobe or the right? Oct 20 '20 at 19:34