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The quadratic $n^2+n+41$ yields prime numbers all the way up to $n=40$ before it fails (pretty cool!).

My question is: Do you know of a quadratic that can 'last even longer'?

Trogdor
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    Note that the Green-Tao theorem tells us that there are arbitrarily long arithmetic progressions in the primes, which would relate to the linear case. – Mark Bennet Nov 17 '14 at 11:53
  • Also, to add to this, is it just a coincidence that the number of positive integer terms that this quadratic lasts HAPPENS to be one less than the constant term? – Trogdor Nov 17 '14 at 11:54
  • Not really a conincidence. It's obvious that $n^2+n+41$ is not prime when $n=41$. – TonyK Nov 17 '14 at 11:55
  • That is true, but unless I am missing something, it is not completely obvious that the prime-producing property necessarily fails precisely at $n-1$. – Trogdor Nov 17 '14 at 11:58
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    $n^2+n+a$ fails to be prime for $n=a-1$: $(a-1)^2+(a-1)+a=a^2$ – Hagen von Eitzen Nov 17 '14 at 12:03
  • I can understand that $n-1$ doesn't work, but my question was more so targeted towards the property that every $n$ less than or equal to $n-2$ generates primes. – Trogdor Nov 17 '14 at 12:05
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    @Trogdor Actually, this is closely related to the fact that $e^{\pi\sqrt{4\cdot 41-1}}$ is very close to an integer! – Hagen von Eitzen Nov 17 '14 at 12:07
  • From where did you manage to pull that expression from? Wow! – Trogdor Nov 17 '14 at 12:09

3 Answers3

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$$36n^2-810n+2753$$

is prime for $n\le44$ (source).

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$$n^2-81n+1681$$ is prime for all natrual numbers $\le 80$. (This is of course an unfair answer)

Hagen von Eitzen
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  • This is the OP's polynomial in disguise, and yields the same primes, though more often :) –  Nov 17 '14 at 12:09
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From Sierpsinski's prime sequence theorem http://mathworld.wolfram.com/SierpinskisPrimeSequenceTheorem.html you can find arbitrarily many primes (but maybe not consecutive values of the quadratic).

Konstantinos Gaitanas
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