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

Considering the equation modulo $3$, we must have $2j^2 \equiv 0 $ which implies that $ j=3k$ for some $k\in \mathbb{N}$.
$ 3i^2 + 18k^2 = 77 \cdot 6^{2012} $ so $$ i^2 + 6k^2 = 77 \cdot 2^{2012} \cdot 3^{2011}.$$
which now implies $i=6l $ for some $l\in \mathbb{N}.$ Thus
$$ 6l^2 + k^2 = 77 \cdot 2^{2011} \cdot 3^{2010}.$$
Similarly, $k=6m$ for some $m\in\mathbb{N} $ and
$$ l^2 + 6m^2 =  77 \cdot 2^{2010} \cdot 3^{2009}.$$
A pattern is emerging that we will keep on having this divisibility by 6 until we reach some $a$ and $b$ such that $$ a^2 + 6b^2 = 77 \cdot 2 =154 $$
which (by checking the small number of possible cases) has solutions $ (a,b) = (10,3)$ and $ (2,5)$ in the naturals, or by varying all possible sign combinations, there are 8 solutions in the integers. Now to find the solutions of the original equation, simply back substitute the factors of 6 we kept extracting. For this problem however that is not required since only the number of solutions is asked, and the answer is 8. 
