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This comes from this question, which I attempted to answer but I got blowback because the original question lacked quality standards. So:

Suppose you are given a proposition such as, "$2x^2+29$ is prime for all (positive) natural numbers $x$ less than or equal to $28$". How can this be proven without directly checking dozens of polynomial values for primality?

Oscar Lanzi
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    But your question is still a PSQ - just like the original - so will also likely be closed as such unless you remedy that. – Bill Dubuque Apr 17 '22 at 13:51

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Let's take a look at $2x^2+29$. This can have a prime factor $p\not\in\{2,29\}$ iff $(29)(-2^{−1})$ is a quadratic residue modulo $p$. Thus the $29$ and $−2$ would have to have the same quadratic character.

Excluding the primes $2$ (which can never divide $2x^2+29$) and $29$ (which is excluded for natural $x≤28$), we require one of the following:

$p∈\{1,3\}\bmod8,p∈\{±1,±4,±5,±6,±7,±9,±13\}\bmod29...(i),$

Or

$p∈\{5,7\}\bmod8,p∈\{±2,±3,±8,±10,±11,±12,±14\}\bmod29...(ii).$

The first of these renders both $−2$ and $29$ as quadratic residues $\bmod p$, the second renders them both nonquadratic residues. The quadratic residues $\bmod 29$, where $29$ is a prime one greater than a multiple of $4$, are obtained from counting out the natural squares until seven nonzero additive inverse pairs $\bmod 29$ are identified.

For $x=28$ our polynomial would equal $1597$, whose square root lies between $37$ and $41$, so if $2x^2+29$ is to be composite for $x≤28$ it must have a prime factor less than or equal to $37$. But:

  • No prime below $31$ satisfies either (i) or (ii). We get a residue $\bmod8$ in the first (quadratic) list but a residue $\bmod29$ in the second (nonquadratic) list, or vice versa.

  • The prime $31$ satisfies (ii), but $31|2x^2+29$ implies $x≡±1\bmod31$, so only $31$ itself (x=1) is in the list for a natural $x≤28$.

  • The prime $37$ satisfies (i), but $37|2x^2+29$ implies $x≡±2\bmod37$, so only $37$ itself (x=2) is in the list for a natural $x≤28$.

We are left with no candidates for composite numbers with $x≤28$.

Oscar Lanzi
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    You are essentially repeating a special case of classical proofs on prime-producing polynomials (but in more elementary language - avoiding basic knowledge of quadratic number fields). – Bill Dubuque Apr 17 '22 at 13:46
  • One thing I later found: by looking at the list of class 2 IQFs I find that $29$ is the largest $n$ for which this works with $2x^2+n$. (No even square-free discriminants beyond $-58$ give class 2.) – Oscar Lanzi Apr 17 '22 at 14:06
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    Oscar, I think you get something with one class per genus, Euler's idoneal numbers. For instance, $6 x^2 + 13$ works. I guess to get a long stretch we need the constant term to be prime. Or $3x^2 + 31 $ with interesting possibles $2x^2 + 2x + 47$ and $6x^2 + 6x + 17$ I know a linear term may be allowed, as the famous $x^2 + x + 41$ nope, (2,2,47) is divisible by 3 every third $x$ But (6,6,17) is pretty good. – Will Jagy Apr 17 '22 at 20:17
  • @willjagy We can tell $2x^2+2x+47$ will be a dud. Complete the square and double to clear fractions --> $(2x+1)^2+93$, which perforce is divisible by $3$ when $2x+1$ is. – Oscar Lanzi Apr 17 '22 at 20:38