$\frac{1}{4} (a^2+ 3 b^2)$ is of the form $(c^2+ 3 d^2)$ If $ 2 \mid (a^2+ 3 b^2)$ and $(a,b)=1$ then $4\mid (a^2+ 3 b^2)$.  How can I show $\frac{1}{4} (a^2+ 3 b^2)$ is also of the form $(c^2+ 3 d^2)$?
Here, clearly $ a$ and $b$ are both odd. 
Let $a=2m+1$ and $ b=2n+1$
$\implies\frac{1}{4} (a^2+ 3 b^2)= m^2 + m+1 +3n^2 +3n$. 
I am stuck here. Can anyone please help how to approach from here. Any help would be appreciated. Thanks in advance.
 A: \begin{eqnarray*}
\frac{(c+3d)^2+3(c-d)^2}{4} =c^2+3d^2.
\end{eqnarray*}
Edit: 
If $ a \equiv b \pmod{4}$ then
\begin{eqnarray*}
\frac{a^2+3b^2}{4} = \left( \frac{a+3b}{4} \right)^2 +3 \left( \frac{a-b}{4} \right)^2.
\end{eqnarray*}
& If $ a \equiv -b \pmod{4}$ then
\begin{eqnarray*}
\frac{a^2+3b^2}{4} = \left( \frac{a-3b}{4} \right)^2 +3 \left( \frac{a+b}{4} \right)^2.
\end{eqnarray*}
and the values in the brackets on the RHS will be whole numbers.
A: $$(x^2+ 3y^2)(z^2 + 3w^2) =(xz - 3wy)^2 +  3(wx+ yz)^2 $$ 
This means that the set of the numbers which can be represented as the form $x^2 + 3y^2$, is closed under multiplication. (This holds for general $x^2 + n y^2$) 
And another observation is, $4 = 1^2 + 3\cdot 1^2$ itself has the representation. 
The situation we have is $xz - 3wy = a$, $wx + yz = b$, $z = 1$, $w = \pm1$. This leads to 
$$ x \mp 3y = a \qquad  \pm x + y = b$$
$$x  = \frac{a+3b}{4}\quad y = \frac{-a+b }{4} \qquad \text{ or}\qquad x = \frac{a - 3b}{4}\quad y = \frac{a+b}{4} $$
For any odd pair $(a, b)$, $x, y$ can be choosen to be integer and satisfies $x^2 + 3y^2 = (a^2 + 3b^2)/4$.
