Is there any common proof which works for all $D\in \{1,2,3\}$ of the following fact mentioned below which is purely elementary?
Let $D\in \{1,2,3\}$. Then any odd prime divisor of $ X^2+ D Y^2$ where $(X,Y)=1$ is again in the same form.
Can we apply some sort of induction to prove this?
Any help will be appreciated. Thanks in advance.
Given Tanner's stipulation, that the prime does not divide either of $x,y,$ the conjecture follows from class number one.
If Legendre symbol $(-4D|q) = -1,$ it is not possible to have $$ x^2 + D y^2 \equiv 0 \pmod q $$ unless both $x,y \equiv 0 \pmod q.$
Special treatment is need for primes dividing $4D.$
If Legendre symbol $(-4D|p) = 1,$ it follows that $p$ is integrally represented by a (reduced) form of that discriminant. For your values of $D,$ the only (reduced) form is the one you named.
You begin to get failures of this, but still well behaved, at class number two. At discriminant $-20,$ the form $2x^2 + 2xy + 3 y^2$ represents primes $7,23$ that are not represented by the principal form $x^2 + 5 y^2.$ But $$ 9^2 + 5 \cdot 4^2 = 161 = 7 \cdot 23 $$