It is known that Euler's polynomials $\,n^2+n+p\,$ ($p\,$ prime) represent a prime for $\,n=0,\,...,\,p-2\,$ if and only if the field $\,Q (\sqrt{1-4p})\,$ has class number $\,h=1$.

The best trinomial of such kind is $\,n^2+n+41$.

But what we can say about quadratic trinomials that generate, for consecutive values of their integer variable, only prime numbers of the form $\,4m+1$?

The best I could find out is the following trinomial: $$p(n)=4\cdot(32\cdot(21-n)-n^2)+1$$ that generates prime numbers of the form $\,4m+1\,$ for $\,n=0,\,...,\,14$.

At the $2$-nd place, I would put the trinomial: $$p(n)=4\cdot(n\cdot(64+n)-1171)+3$$ that generates (eventually in absolute value) prime numbers of the form $\,4m+1\,$ for $\,n=1,\,...,\,14\,$ and of the form $\,4m+3\,$ for $\,n=15,\,...,\,28$.

Every suggestion is well accepted.

Many thanks.

[ This question have been also posted on MathStackExchange ]

  • $\begingroup$ Using an extension of Green-Tao Theorem to polynomial patterns (see here) you can show that there are trinomials which express arbitrarily long strings of primes. I suspect a generalization to primes in a fixed arithmetic progression would be immediate from the method of proof, like for standard Green-Tao, but I can't speak with confidence. $\endgroup$ – Wojowu Mar 29 at 11:21
  • 1
    $\begingroup$ Crossposted at MSE. $\endgroup$ – Dietrich Burde Mar 29 at 11:40

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