Are there infinitely many prime numbers of the form $p^{2h}+p^h+1$, where $p$ is a prime and $h$ is a positive integer?
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Already $p^2+p+1$ should be prime for infinite many primes $p$ , although we cannot prove it. The same for $p^6+p^3+1$ – Peter Nov 25 '20 at 18:49
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Aren’t they divisible by $3$ if $p\ne3$? – J. W. Tanner Nov 25 '20 at 18:52
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1@J.W.Tanner Not if $p \equiv 2 \mod 3$. – Robert Israel Nov 25 '20 at 18:55
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For primes $p$ such that $p^2+p+1$ is prime, see OEIS sequence A053182. For primes $p$ such that $p^6+p^3+1$ is prime, see A066100. – Robert Israel Nov 25 '20 at 19:02
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Despite the evidence for infinite many such primes , it is probably out of reach to prove it. – Peter Nov 25 '20 at 19:07
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@RobertIsrael: you’re right, and if $h\not\equiv0\pmod3$ – J. W. Tanner Nov 25 '20 at 19:18
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1Indeed, there are no integer polynomials of degree $>1$ that have been proven to generate infinitely many primes. – Robert Israel Nov 25 '20 at 19:18
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@RobertIsrael OP didn't say they are generated by the same polynomial – Raffaele Nov 25 '20 at 23:02
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@Raffaele Actually, $p^{2h}+p^h+1 = x^2 + x + 1$ where $x = p^h$, so if the answer to OP's question is yes, there are infinitely many primes generated by the polynomial $x^2+x+1$ for integer $x$. – Robert Israel Nov 26 '20 at 04:14
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@J.W.Tanner No, you mean $h \not\equiv 0 \pmod 2$. Which is a good thing, since as shown in my answer $h$ must be a power of $3$. – Robert Israel Nov 26 '20 at 04:18
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@RobertIsrael: you’re right again – J. W. Tanner Nov 26 '20 at 04:50
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The generalized Bunyakovsky conjecture implies that there are, as long as $x^{2h} + x^h + 1$ is irreducible (which I think is true if and only if $h$ is a power of $3$).
EDIT: Yes, that is the case. $x^{2h}+x^h+1 = \dfrac{x^{3h}-1}{x^h - 1}$ is the product of the cyclotomic polynomials $C_d(x)$ where $d$ divides $3h$ but not $h$. If $h$ is a power of $3$, the only such $d$ is $3h$ itself: $x^{2h} + x^h + 1 = C_{3h}(x)$ is irreducible. If $h$ is divisible by some prime $q \ne 3$, then $d = 3h/q$ is another $d$, and $x^{2h}+x^h+1$ is reducible.
Robert Israel
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