When $n>3$, $n$ must be $\equiv 0, 2 \pmod 6$
$3$, $7$, $43$, $73$, $157$, $211$, and $421$ are primes, but $343=7^3$ ($n=18$) is not.
Are there infinitely many primes of the form $n^2+n+1$ ?
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1What makes you think that the answer to this question is known? – John Gowers Jan 23 '18 at 15:11
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If you'd like to know, an affirmative answer to this problem would lead to a breakthrough – ArtW Jan 23 '18 at 15:13
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by accident, I was observing uniqueness of positional numeral systems which all digits are same (likewise 111,444,666...) – Jay C. Lee Jan 23 '18 at 15:18
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Is this an open problem? – Jay C. Lee Jan 23 '18 at 15:20
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5As far as I know there isn't a single example of a non-linear polynomial which is known to take on infinitely many prime values. Of course, there are lots of conjectures regarding this issue, notably The Bunyakovsky conjecture. – lulu Jan 23 '18 at 15:22
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2Yes, it is a special case of the Bunyakovsky conjecture, see this similar question. – Dietrich Burde Jan 23 '18 at 15:23
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incredibly difficult... it's scaring me – Jay C. Lee Jan 23 '18 at 15:32
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@JayC.Lee The problem is open, but very likely the answer is that there are infinite many such primes. – Peter Jan 23 '18 at 18:41
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similar to your polynomial are these two polynomials which also produce primes $p(n)= n^2 + n - 1$ and $p(n)= n^2 - n + 1$ I noticed that the first one produces more primes but it may change if we test larger and larger number $n$. – user25406 Jan 24 '18 at 00:32