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According to Lemke-Oliver, irreducible quadratic polynomials $G$ with positive leading coefficient and $\rho(2)<2$, (where $\rho(m)$ denotes the number of incongruent solutions to the congruence $G(n) \equiv 0\ (\mathrm{mod}\ m)$) represent $p_1p_2$ infinitely often, with $p_1,p_2$ two distinct primes.

Is there anything known about the distribution of $(p_1,p_2)$, with $(\cdot,\cdot)$ the Legendre symbol? I am working with a specific family of polynomials which appear to have $(p_1,p_2)=\pm1$, with each case occurring infinitely often (and with the same density). Is there any known results along these lines?

Update: if this is unknown, can we at least argue that the density of $(p_1,p_2)=\pm1$ are both strictly positive? That is, that there is no polynomial where only one of the options is realised. Or is this out of reach too?

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    $\begingroup$ The short answer is that probably nothing is known, but your expectation should be correct. One of the issues is that the current way to prove that a quadratic $f$ represents semi-primes infinitely often uses a sieve, which doesn't quite pick up the correct order of magnitude of semi-primes of order 2. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 12, 2018 at 1:28
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    $\begingroup$ I unfortunately have to agree with Stanley that this is hard. In fact, though it's been a while since I've thought about the problem, I believe I only proved that there are infinitely many values G(n) with at most two prime factors, not exactly two, so if you really need semiprimes, you might be out of luck. OTOH, if you want something weaker (e.g., that there's some factorization of G(n) = d_1*d_2, not necessarily into primes, with (d_1,d_2) of each sign), then there might be some games you could play, but this would have very a different feel. $\endgroup$
    – rlo
    Commented Nov 13, 2018 at 18:52
  • $\begingroup$ @rlo Oh, it's great to have your input here! Thank you for your comment, it is very useful (indeed, I thought that $P_2$ were numbers with exactly two prime factors, not up to; my bad). In my case, prime or semi-prime are both fine, but other composites do not really work. In fact, the existence of a single prime, or a single pair with $(p_1,p_2)=-1$ would suffice, but I guess that's too much to ask... Anyway, thank you again! $\endgroup$
    – Delmastro
    Commented Nov 13, 2018 at 22:28

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