$\begingroup$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.
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$\endgroup$41 Answer
$\begingroup$$$f=36x^2 - 192x + 433$$ produces only integers of the form $\ 4k+1\ $ and for integers in then range $\ [0,36]\ $, it is only composite for $\ 28\ $ and $\ 29\ $. In particular, it is prime for integer $\ x\ $ with $\ 0\le x\le 27\ $
I restricted the first two coefficients to be divisible by $\ 4\ $ and the last being of the form $\ 4k+1\ $, which ensures that only values of the form $\ 4k+1\ $ can occur. With those restrictions and with coefficients with absolute value not larger than $\ 600\ $, this is the best possible polynomial.
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