Number of integer solutions to $3i^2 + 2j^2 = 77 \cdot 6^{2012}$ here is another problem I did not manage to solve in the contest I mentioned in my previous question.

Determine the number of integer solutions $(i, j)$ of the equation:
  $3i^2 + 2j^2 = 77 \cdot 6^{2012}$.

Applying logarithms is not useful, since on the left hand side we have a sum; I also tried some algebraic manipulations that led me to nothing useful. Is there a simple solution to the problem?  
Thank you,
rubik

$A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$
$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :
$i= \pm 5x \Rightarrow i = \pm 5 \cdot 6^{1006}$
$j= \pm x \Rightarrow j = \pm  6^{1006}$
$R_A :$ 
$(i,j) \in \{(-5 \cdot 6^{1006},-6^{1006}),(-5 \cdot 6^{1006},6^{1006}),(5 \cdot 6^{1006},-6^{1006}),(5 \cdot 6^{1006},6^{1006})\}$

$B)~ i^2=9x^2 ~\text {and}~ j^2=25x^2$
$3 \cdot 9x^2+2 \cdot 25x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :
$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$
$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$
$R_B :$
$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$
