Can we prove that certain quadratic polynomials generate a long list of primes without trial and error? 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?
 A: 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 $x≤28$.
