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: 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 !
A: 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. 
A: I find it easiest to proceed as follows. First note that $j$ must be divisible by $3$ and $i$ must be even. But then $3i^2$ is divisible by $9,$ so $i$ is divisible by $3,$ and similarly $j$ is even. Suppose then that $6^k$ divides ${\rm gcd}(i,j)$ but $6^{k+1}$ does not. Then $ 1 \leq k \leq 1006.$ Write $i = 6^{k}s$ and $j = 6^k t.$ If $k < 1006,$ we have $36$ divides $3s^2 + 2t^2,$ and again $3$ divides both $s$ and $t$, and $s$ and $t$ are both even, contadicting the choice of $k,$ since $6$ then divides ${\rm gcd}(s,t).$ Hence we are reduced to finding the number of integer solutions to $3s^2 + 2t^2 = 77.$ Now $s$ must be odd, so $3s^2 \equiv 3$ (mod $8$) and hence $2t^2 \equiv 2$ (mod $8$), and $t$ is also odd. Thus $s$ an $t$ are both odd. Clearly $|s| \leq 5.$ If $|s| = 5,$ then $|t| = 1,$ while if $|s| = 3$ then $|t| = 5.$ If $|s| = 1,$ then there is no solution for $t.$ Hence the solutions for $s$ and $t$ are $(5,1),(5,-1),-5,1),(-5,-1), (3,5),(3,-5),(-3,5),(-3,-5),$ and each solution for $(s,t)$ yields the unique solution $6^{1006}(s,t)$ for $(i,j).$
A: Hint $\ \ \ $ Prime $\rm\:p\nmid a,\:\ p^2\ |\ n = a\: x^2 + p\:y^2\:\Rightarrow\: p\ |\ x^2\:\Rightarrow\: p\ |\ x\:\Rightarrow\: p\ |\ y^2\:\Rightarrow\: p\ |\ y $  
Therefore, we deduce  $\rm\:\ n/p^2 = a\:\bar x^2 + p\:\bar y^2,\ $ for $\rm\ \bar x = x/p,\ \bar y = y/p \in\mathbb Z$ 
So for $\rm\: p = 3,2,\ \ \ 3\:i^2 + 2\:j^2 = 77 \cdot 6^{2012}\:\Rightarrow\: 3\:\bar i^2 + 2\:\bar j^2 = 77\ $ for $\rm \bar i = i/6^{1006},\:\bar j = j/6^{1006}\in\mathbb Z$
A: $A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$
$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , 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 :
$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$
$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$
$R_B :$
$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$
