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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.

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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

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    $\begingroup$ 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 $\endgroup$
    – R-B
    Oct 20 '20 at 18:57
  • $\begingroup$ @R-B No, it starts with $n,n+1,n+2,\ldots,$, right? We have $k!+3=(k!+2)+1=n+1$. $\endgroup$ Oct 20 '20 at 21:59
  • $\begingroup$ But yes, you are right. There is also a nice solution using CRT, see the duplicate Bill pointed out.. $\endgroup$ Oct 21 '20 at 8:16
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Hint : Try to prove that for all $k\geq 0$, $$n=(1002+k)!+2$$

satisfy the property.

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

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