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 MarinoCommented Dec 19, 2023 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 AsimovCommented Dec 19, 2023 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, 2023 at 11:51
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$\begingroup$ More generally, if $f(x) \in \mathbf Z[x]$ is nonconstant and $c \geq 1$, then there is prime $p$ s.t. $f(p)$ has at least $c$ prime factors. Passing to an irred. factor of $f(x)$, we may assume $f(x)$ is irred. Let $\alpha$ be a root of $f(x)$. By algebraic number theory, infinitely many primes split completely in $\mathbf Q(\alpha)$ and don't divide ${\rm disc}(f)$ or the leading coeff. of $f(x)$, so we can solve $f(r_i) \equiv 0 \bmod q_i$ for all $i$. When $i = 1,\ldots,c$, there's a prime $p$ s.t. $p \equiv r_i \bmod q_i$ by Dirichlet, so $f(p) \equiv 0 \bmod q_i$ when $i=1,\ldots,c$. $\endgroup$– KConradCommented Dec 27, 2023 at 17:51
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$\begingroup$ The usual proof that each number field has infinitely many primes split completely in it uses properties of zeta-functions of number fields. But there is a simpler argument, which was given by Bjorn Poonen on MO in 2010 (every number field is in a Galois extension of $\mathbf Q$, so it suffices to do this with Galois extensions, which is what Bjorn does): mathoverflow.net/questions/15220/… $\endgroup$– KConradCommented Dec 27, 2023 at 17:57
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$\begingroup$ Keith, do we really need algebraic number theory here, or even passing to an ireducible factor? All steps seem to be quite easily modified for arbitrary non-constant polynomial with integer coefficients. $\endgroup$ Commented Dec 27, 2023 at 18:14
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$\begingroup$ You're right, the argument doesn't need to use primes that split completely in a number field. And it has already appeared on MSE: math.stackexchange.com/questions/1019538/… $\endgroup$– KConradCommented Dec 27, 2023 at 18:29
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$\begingroup$ Of course this is well known. Another type of argument is: assume the contrary: all divisors of $f(n)$ are bounded by $C$, i.e., all values of $f$ at integer points are $C$-smooth, but $C$-smooth numbers are much more rare than the values of a polynomial (some power of $\log N$ vs. $\sqrt[d]N$ numbers not exceeding $N$, respectively, where $d=\deg f$). $\endgroup$ Commented Dec 27, 2023 at 20:07
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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.
answered Dec 19, 2023 at 14:07