Is it true that, for each positive integer $c$, there exists a prime number $p$ such that $p^2+p+1$ is divisible by at least $c$ distinct primes?
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3$\begingroup$ Just a small comment: $p^2+p+1\equiv 0 \pmod{q}$ reformulates as $2p+1 \equiv \pm \sqrt{-3} \pmod{q}$. By reciprocity law, $-3$ is quadratic residue iff $q \equiv 1 \pmod{3}$ (or $q=3$, which is irrelevant if we want many primes). Another consequence of reciprocity is that $p \equiv (p+1)^2 \pmod{q}$ , implying that $q$ has fixed quadratic residue behaviour (depending on $(-1)^{\frac{(p-1)(q-1)}{4}}$) $\endgroup$– Andrea MarinoDec 19 at 11:25
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1$\begingroup$ Since p^2 + p + 1 is the size of the projective plane of the prime field F_p, I wonder if there is a connection to finite projective planes. $\endgroup$– Daniel AsimovDec 19 at 19:32
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2 Answers
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Yes. At first, there exist $c$ distinct primes $q_1,...,q_c$ which divide some $m_i^2+m_i+1$ for $i=1,\ldots,c$ respectively (induction on $c$: if you found $c-1$ such primes, take $m_c$ being equal to their product and $q_c$ equal to arbitrary prime divisor of $m_c^2+m_c+1$). Next, by Chinese remainder theorem there exist $m$ congruent to $m_i$ modulo $q_i$ for $i=1,\ldots,c$. Finally, by Dirichlet theorem there exists a prime $p$ congruent to $m$ modulo $\prod q_i$.
answered Dec 19 at 11:51
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There is the following theorem of Halberstam, "On the distribution of additive number-theoretic functions. III." Let $\omega(n)$ be the number of prime factors of $n$. Given any irreducible polynomial $f(x)$ with integer coefficients, we have $$\sum_{p\leqslant n} \omega(f(p))> \frac{cn\log\log n}{\log n}$$ with $c>0$. Also, for all but $o(n/\log n)$ primes $p\leqslant n$, $\omega(f(p))=(1+o(1)) \log\log n$. The proof is not straightforward, and uses some deep results on primes in arithmetic progressions.
Applying this with $f(x)=x^2+x+1$ gives you a quantitative version of what you want.
