# A quadratic trinomial that generates only prime numbers of the form $4m+1$

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 ]

• 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. – Wojowu Mar 29 at 11:21
• Crossposted at MSE. – Dietrich Burde Mar 29 at 11:40