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It is well-known that a prime number $p$ is $\equiv 1 \pmod 4$ iff $p=x^2+y^2$ for some integers $x,y$ (except for $p=2$). My question is: is there an irreducible homogeneous polynomial $f \in \Bbb Z[x_1, ..., x_n]$ of total degree $>1$, such that $p \equiv 3 \pmod 4$ iff $p \in \mathrm{Im}(f)$ (up to finitely many exceptions) ? I saw this question, but this is also the condition $p \mid f(y)$ for some $y \in \Bbb Z^n$ (and not $p=f(y)$ as I want).

More generally, given $M \geq 1$ and $S \subset \Bbb Z/M\Bbb Z$, when is there $n \geq 1$ and an irreducible homogeneous polynomial $f \in \Bbb Z[x_1, ..., x_n]$ of total degree $>1$, such that a prime $p$ is in the image of $f$ iff $[p]_M \in S$, up to finitely many exceptions?

Alphonse
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    @JyrkiLahtonen : if $p = 3 \pmod 4$, then $w := p-2 = 1 \pmod 4$, but $w$ might not be prime, so we can't deduce that $w=x^2+y^2$ for some $x,y$ (it depends on the various $\ell$-adic valuations of $w$…) – Alphonse Jun 07 '18 at 09:24
  • Oops. That was silly of me. Sorry. – Jyrki Lahtonen Jun 07 '18 at 09:25
  • @JyrkiLahtonen : it was not silly at all! I first thought that your idea worked (and I felt silly). Probably other people will have this idea. – Alphonse Jun 07 '18 at 09:28
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    Anyway, I think that in the case of binary quadratic forms this question is well understood. Many cite Cox's book. IIRC the forms sharing the same discriminant cover primes in different residue classes modulo the discriminant, but they do not cover all the coprime residue classes. As you observec, the discriminant $\Delta=4$ is a case in point. Algebraic number theory (and class field theory) give answers as to which residue classes are covered. If Will Jagy (or somebody else in the know) shows up... – Jyrki Lahtonen Jun 08 '18 at 09:28
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    With higher degree forms - it's complicated. Non-abelian class field theory won't spew out residue classes (but I don't know the details). From cyclotomic extensions we may get something more, but a prime is presented by the norm polynomial iff it splits completely, when it will also split in the quadratic intermediate field... Sorry, entering a territory where I may unwittingly give misinformation :-( – Jyrki Lahtonen Jun 08 '18 at 09:35
  • Can't you just pick $4x_1+3\in \mathbb{Z}[x_1]$ or am I missing something here? – Mastrem Jun 08 '18 at 17:36
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    @Mastrem, yes, that is indeed correct, so now the OP has changed the question to add a requirement on the degree. I'll offer the polynomial $(4x_1 + 3)(x^2_2 + x^2_3 + x^2_4 + x^2_5)$ so the OP can change the question again. – Infinity Jun 08 '18 at 19:23
  • @Infinity : thank you. I hope it is a correct version now. – Alphonse Jun 08 '18 at 21:21
  • Apparently you missed the point of my comment, which is that changing the question is poor form. You are not "correcting" the question, you are just moving the goalposts. – Infinity Jun 09 '18 at 00:33
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    Oh, and $4(x^2_1 + x^2_2 + x^2_3 + x^2_4) + 3$. – Infinity Jun 09 '18 at 00:34
  • And... answer to version #four of this question is $4(x^2_1 + x^2_2 + x^2_3 + x^2_4) + 3 x^2_5$. I'm not going to bother looking at any further modifications to this question (for the reader, the current version is homogenous, irreducible, degree (at least 2), forms, the three adjectives in reverse order corresponding to the second, third, and fourth version of this question. – Infinity Jun 09 '18 at 19:03

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