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

Observe that $i$ has to be even, so do $j$ because $3i^2$ and the right side can be divided by 4.
Let $i'=2i$, $j'=2j$; the equation becomes $3i'^2+2j'2=77 \cdot 3^{2012} \cdot 2^{2010}$. The same argument applies 1006 times, and we get something like $3i^2+2j^2=77 \cdot 3^{2012}$.
Now do the same looking the divisibility by 3. It works. Finaly check $3i^2+2j^2=77$
Edit: It's my first posting here, so hello to all from France !
