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Prove that for every integer $a, b$ there exists an integer $n$ so that the number $n^2+an+b$ has at least 2018 prime divisors

I tried:

  • Factoring the number $n^2+an+b = n(n+a) +b$
  • Considering the sequence ${P_n}$ of all primes in increasing order
Math Buster
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    Sounds like a fun problem. But I need to ask whether this might be from some contest? You see, contest questions often use the year as an input in question (sometimes the exact year is crucial, sometimes a mild congruence condition would be ok, sometimes the parameter is a red herring). AND we have a strict policy not to allow questions from on-going contests. – Jyrki Lahtonen Jul 03 '18 at 19:06
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    If not from a contest then you probably should give a bit of other context. What pieces of theory have been covered recently? Chinese remainder theorem? Quadratic residues? Reciprocity law? – Jyrki Lahtonen Jul 03 '18 at 19:09
  • In questions like this I first use the brute force solution to get a feel for the problem and then try to derive a more efficient solution. – Weather Vane Jul 03 '18 at 19:12
  • The contest has already ended, I was not part of it. But my friends wouldn't let me know the answer to it. All I know is I can use ALL arithmetic theorems – Math Buster Jul 03 '18 at 19:12
  • Do you know the Chinese remainder theorem? – saulspatz Jul 03 '18 at 19:13
  • @saulspatz yes, I know the Chinese remainder theorem. – Math Buster Jul 03 '18 at 19:16
  • @MathBuster I was in the middle of typing a hint on how to use the Chinese remainder theorem when an answer was posted, so I abandoned the hint. – saulspatz Jul 03 '18 at 19:32

1 Answers1

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Assuming that $p$ is an odd prime, the equation $$ n^2+an+b \equiv 0\pmod{p} $$ has at least a solution $n\equiv c_p\pmod{p}$ as soon as $a^2-4b$ is a quadratic residue $\!\!\pmod{p}$.
$a^2-4b$ is a quadratic residue for infinite$^{(*)}$ primes $p_1,p_2,p_3,\ldots$ and the system $$ n\equiv c_{p_1}\!\!\!\pmod{p_1},\quad \ldots,\quad n\equiv c_{p_{2018}}\!\!\!\pmod{p_{2018}} $$ has an integer solution by the Chinese remainder theorem.
A slight generalization is that $\omega(\text{monic quadratic polynomial }(n))$ is unbounded.

$(*)$ This is not entirely trivial. Assume, by contradiction, that some integer $m$ is a quadratic residue only for a finite number of prime moduli, the largest of them being $p$. By Dirichlet's theorem there is a prime $Q\equiv 1\pmod{4m}$ such that $Q>p$. By quadratic reciprocity for Legendre/Jacobi symbols $$ \left(\frac{m}{Q}\right)=\left(\frac{Q}{m}\right)=\left(\frac{1}{m}\right)=1$$ hence $m$ is a quadratic residue for some prime $Q>p$, contradiction.

Jack D'Aurizio
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