Form of divisors of Euler's prime generating polynomial I found an interesting form of divisors of Euler's prime generating polynomial.
The form is this.
$$
x^2+xy+41y^2
$$
This expression exactly become to Euler's prime generating polynomial when $y=1$.
Do you know something about this form ?
 A: There are two related facts, tied together because of class number one:
if we have odd prime $q$  with Legendre symbol $(-163|q) = -1,$   then
$$ x^2 + xy + 41 y^2 \equiv 0 \pmod q  $$
if and only if
$$  x \equiv y \equiv 0 \pmod q,$$
in which case $x^2 + xy + 41 y^2$ is actually divisible by $q^2.$ The same fact holds for the prime $2$ because $41$ is odd, the form is not even unless both $x,y$ are even.
Next, if you have an odd prime $p$ dividing $x^2 + x + 41,$ it is necessary that either $p=41$ or $(-163|p) = 1.$ This comes from taking $y=1$ in the first paragraph. Oh, of course $x^2 + x + 41$ is odd.
Finally, if $(-163|p) = 1$ or $p=41,$ then $p$ really can be expressed as
$$ p = u^2 + uv + 41 v^2.  $$ That also comes of class number one. There is a simple algorithm based on first solving
$$  \beta^2 \equiv -163 \pmod{4p}$$
or $\beta^2 - 4pt = -163,$ then keeping careful track while Gauss reducing the positive binary form $\langle p, \beta, t \rangle$  to the inevitable $\langle 1,1,41 \rangle$ and inverting a certain 2 by 2 integer matrix.
It was an elementary result of Rabinowitz in 1913 that the polynomial $x^2 + x + p$ with $p$ a positive odd prime, and form class number one,  assumes prime values with all integers $0 \leq x \leq p-2.$ I put a proof at
Is the notorious $n^2 + n + 41$ prime generator the last of its type?
It is actually an if and only if result.
