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 $i&#39;=2i$, $j&#39;=2j$; the equation becomes $3i&#39;^2+2j&#39;2=77 \cdot 3^{2012} \cdot 2^{2010}$. 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 !
$A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$

$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , 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 :

$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$

$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$

$R_B :$

$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$