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Show that $2n^2+11$ and $2n^2+29$ generate primes for all non-negative integers $n<11$ and $n<29$ respectively.

This question is from Elementary Number Theory by Kenneth Rosen. The book's solution has used a brute force method to generate all numbers and manually check if they are prime or not.

I wish to ask instead if there is a better method. I have looked around on the internet and couldn't find anything. I instead wrote a C program to test the theory about primes of the form $2n^2+x$ $(0leq n<x)$, and it seems only $x=3,5,11,29$ satisfy it under $10^6$.

So, is there no better, more number theoretical way to verify the given question?

Context: here is a proof for the better-known $n^2 + n + 41$ Is the notorious $n^2 + n + 41$ prime generator the last of its type?

Took me a while and some numerical experiments, but I got this one. For foundational material on positive binary quadratic forms, see https://en.wikipedia.org/wiki/Binary_quadratic_form and https://en.wikipedia.org/wiki/Binary_quadratic_form#References My favorite book is Buell; people seem to like Cox.

We are given a prime $p equiv pm 3 pmod 8$ such that the form class number $h(-8p) = 2.$ In particular, the only classes are those of $$ x^2 + 2py^2 $$ and $$ 2 x^2 + p y^2. $$ The shorthand for these two forms is $langle 1,0,2p rangle$ and $langle 2,0,p rangle ; . ;$

Note that the product $2p$ must be one of Euler's Idoneal Numbers

We will ASSUME that there is an integer $n$ with $1 leq n < p$ such that $2n^2 + p$ is composite. We will show that this leads o a contradiction by constructing a third primitive quadratic form of the same discriminant.

Let $q$ be the smallest prime that divides any $2n^2 + p$ with $n < p.$ Usually there will be several small $n$ giving the same $q.$ In turn, take the smallest $n$ that works, meaning $2n^2 + p equiv 0 pmod q ; . ;$ We get $q$ prime,
$$ 2n^2 + p = q t ; , ; $$
$$ t > q > 2n ; . ; $$

There are just two cases.

IF $q > 4 n,$ then
$$ langle q,4n,2t rangle $$
is a reduced primitive form of the same discriminant $(-8p).$ It is not equivalent to either of the original forms listed, that is what we get for reduced forms.

IF $q < 4 n,$ then
$$ langle q,4n-2q, 2t - 4n + q rangle $$
is a reduced primitive form of the same discriminant $(-8p).$

In either case, the presence of composite numbers represented as $2n^2 + p$ has resulted in $h(-8p) geq 3,$ contradicting the hypothesis of class number 2.

Examples for the first case:

If $p equiv 3 pmod 5,$ we can take $n=1$ and get third form
$$ langle 5,4, frac{2p+4}{5} rangle $$

If $p equiv 3 pmod {11},$ we can take $n=2$ and get third form
$$ langle 11,8, frac{2p+16}{11} rangle $$

Examples for the second case:

If $p equiv 1 pmod 3,$ we can take $n=1$ and get third form
$$ langle 3,2, frac{2p+1}{3} rangle $$
If $p equiv 2 pmod 5,$ we can take $n=2$ and get third form
$$ langle 5,2, frac{2p+1}{5} rangle $$

Alright, I will need to work on this a bit, but it is already clear that your condition is that (binary quadratic) form class number $h(-8p) = 2$ for prime $p.$ Note that your $p equiv pm 3 pmod 8,$ otherwise the form $langle 2,0,p rangle$ would be in the principal genus, but the presence of two ambiguous forms demands a second genus, therefore the class number would be at least four.

Discr  -24 = 2^3 * 3  class  number  2



 all  

      24:  < 1, 0, 6>    Square        24:  < 1, 0, 6>

      24:  < 2, 0, 3>    Square        24:  < 1, 0, 6>





Discr  -40 = 2^3 * 5  class  number  2



 all  

      40:  < 1, 0, 10>    Square        40:  < 1, 0, 10>

      40:  < 2, 0, 5>    Square        40:  < 1, 0, 10>





Discr  -88 = 2^3 * 11  class  number  2



 all  

      88:  < 1, 0, 22>    Square        88:  < 1, 0, 22>

      88:  < 2, 0, 11>    Square        88:  < 1, 0, 22>





Discr  -232 = 2^3 * 29  class  number  2



 all  

     232:  < 1, 0, 58>    Square       232:  < 1, 0, 58>

     232:  < 2, 0, 29>    Square       232:  < 1, 0, 58>