<s>[INST] If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.
I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.
I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?
EDIT: Sorry, here is the "closed under multiplication" proof I found:
Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.
EDIT (2): I am very sorry. The question should be:
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
 [/INST] This is false as stated. $4$ has an expression as $a^2 + 3 b^2,$ with $a=b=1,$ but $2$ does not. $25$ has such an expression, with $a=5,b=0,$ but $5$ itself does not. The same applies to any $q^2,$ where $q \equiv 2 \pmod 3$ is prime. 
</s>