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Question: Let $n \in \mathbb{Z}$ where $n \geq 2$, prove no number in the list: $$n! + 2, n! + 3,...,n! + n$$ is prime.

I have written my proof exactly as follows:

Proof: $P(n) = n! + n = n((n-1)! + 1)$. Therefore both $n$ and $((n-1)! + 1)$ are factors, and therefore no number $p(n),n \geq 2$ is prime.

Is a proof of this form acceptable? Have I missed anything?

In my understand a proof should simply look like:

Statement->Proof->Conclusion

Is this correct?

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Also quick question, how do you place some of your posts text in a darkened square? How do you place separation lines around your proof for example? –  Display Name Mar 16 at 2:45
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You can put > in front of a line to make it appear in a block quote. If you need a horizontal rule, you can put ***. I've formatted your post a little more. –  George V. Williams Mar 16 at 2:50

3 Answers 3

up vote 4 down vote accepted

Your proof is not acceptable as stated: you need to point out that neither $n$ nor $(n-1)!+1$ are 1. (This is easy, of course; you stated $n>2$ so the first term is not one, and $(n-1)!>0$ so $(n-1)!+1>1$.) As mentioned in breeden's answer, you haven't proved the theorem, either. You need to show

$n!+k$ is not prime for $2 \leq k \leq n$.

The argument proceeds similarly to before... instead of factoring out an $n$, you need to factor out a $k$.

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I think I may not understand the list fully: Say I choose n = 5, will a list be generated that is $5! + 2, 5! +3, 5! + 4, 5! + 5$? Instead of what I assumed: $\forall n \in \mathbb{Z}, n\geq 2$ -> $2!+2,3!+3,4!+4,...,n! + n$ –  Display Name Mar 16 at 3:25
    
Look at the statement of your problem: it says that it's your first list that will be generated, not the second. –  Mike Miller Mar 16 at 3:30
    
I see now, thank you. I suppose I just need to expose myself to more of these problems(Need Mathematical maturity...). Thanks for the clarification –  Display Name Mar 16 at 3:32
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@DisplayName The key is to read the statements of problems carefully - I can't tell you the number of times something has been too easy or too hard because I didn't read the problem right! –  Mike Miller Mar 16 at 3:33

You only proved that one number in the list is not prime, in particular the last number $n! + n$. What about the other $n - 2$ numbers: $n! + 2, n! + 3, \dots, n! + (n - 1)$?

Though your proof that the $n! + n$ is not prime is acceptable. Following the Statement -> Proof -> Conclusion format is recommended, though what is actually required depends on your audience (in your case, your professor and you!).

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So I need to inductively show that the list is equivalent to my direct form? –  Display Name Mar 16 at 2:55
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@DisplayName No, your form is NOT equivalent to the list. –  Mike Miller Mar 16 at 2:56

If $ 2 \leq k \leq n $: $$ k < n!+k=k(\frac{n!}{k}+1) \implies k|(n!+k)$$

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