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I'm having difficulty proving this statement.

For every positive integer $n$, $n^2-n+11$ is a prime

So I know the obvious, that a prime can only be divided by $1$ and itself, but I'm not sure how to turn that into some generic equation to compare to the one mentioned in the question. Could someone give me a start?

ADH
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    Hint: consider the case of $n=11$. – Regret Mar 10 '15 at 20:17
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    Ah, so a counter example to disprove it? – ADH Mar 10 '15 at 20:18
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    Yes, precisely! – Regret Mar 10 '15 at 20:18
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    Duh, thanks a lot! – ADH Mar 10 '15 at 20:18
  • Not that it matters, but if you need them, I have plenty more counterexamples where that one came from ... – David K Mar 10 '15 at 20:19
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    Just for your information, there are no closed formula functions in general that are prime for all $n$. – Gregory Grant Mar 10 '15 at 20:23
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    @Gregory: What about the constant function $2$? – Regret Mar 10 '15 at 20:26
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    Don't be smart, lol. You know what I meant. More precisely, there are no non-constant closed formula functions $f:\mathbb N\rightarrow\mathbb N$ all of whose values are prime s.t. $|f(\mathbb N)|=\infty$. – Gregory Grant Mar 10 '15 at 20:30
  • @Gregory: Hehe. Sorry, I could not resist. I suppose that depends on your definition of closed form, doesn't it? Does Mills' formula count as closed form? – Regret Mar 10 '15 at 20:33
  • @ADH Note that a number is prime if and only if it has exactly two distincts divisors. In particular $1$ is not prime although it is only divisible by $1$ and itself. – Surb Mar 10 '15 at 20:37
  • @Regret, Mill's would be if we could know what $A$ is without first finding all primes in the first place. I should have said no "known" formula because I don't think it has been proven that no such formula can exist in principle. – Gregory Grant Mar 10 '15 at 20:39
  • @Surb such distinction is really unnecessary, since it isn't meaningful to talk about a "number of divisors" at all, if we are not talking about distinct divisors. (If we aren't talking about "distinct divisors", then any number has infinite divisors. It can be taken for granted, by virtue of the fact that we are specifying number in the first place.) – Jonathan Hebert Mar 10 '15 at 20:42

2 Answers2

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Hint $\ f(11n)\, =\, 11\,(11n^2\!-n+1)\,$ always prime implies $\,11n^2\!-n+1 = 1 $ for all $\,n>0,\,$ contra a nonzero quadratic has at most $2$ roots.

Remark $\ $ I presented the proof this way because exactly the same proof shows that any nonconstant polynomials cannot produce only primes. You may find it much more instructive to prove this more general statement.

Community
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Bill Dubuque
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let $n=11$

$n^2$ is a multiple of $11$.

A multiple of $11$ minus $n$(a multiple of 11) is also a multiple of 11.

When a multiple of 11 is added to a multiple of 11, it is again a multiple of 11.

It means, 11 is a factor of the number -> the number has a proper factor 11.

Therefore $n$ is not prime.

Especially Lime
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Sriniketh
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