$\frac{1}{4} (a^2+ 3 b^2)$ 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.

\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.
