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I wrote a program to check and found, when $0\leq n \leq 39$, $n^2+n+41$ are primes.

But now I wonder: does it have a simple method to prove when $0\leq n \leq 39$, $n^2+n+41$ are primes without check it one by one?

Update: A similar question Prove that $n^2+n+41$ is prime for $n<40$

Community
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    I put an elementary proof of Rabinowitz(1913) at http://math.stackexchange.com/questions/289338/is-the-notorious-n2-n-41-prime-generator-the-last-of-its-type/289357#289357 – Will Jagy Sep 06 '15 at 01:55

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http://planetmath.org/proofthatn2n41isprimefor0leqnleq40

I hope that this will help you, a little bit at least.

Curious me
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  • The planetmath proof is very unsatisfactory, given the OP's request. I suspect there is a truly elementary proof based on a clever partition of $41$ and an associated inequality. – guest Sep 06 '15 at 01:53
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    @guest, the original proof, both directions, by Rabinowitz in 1913 amounts to this, although he wrote in number field language. I wrote both directions in quadratic forms style, see my comment directly below the question. – Will Jagy Sep 06 '15 at 02:04
  • @WillJagy disregard my last comment, I misunderstood. So there is no proof for the particular form that does not involve equivalence of forms? – guest Sep 06 '15 at 03:00