The equation is equivalent to :$9[(x+y)^2-2xy+1]+2(3xy+2)=2005$ Now we put $x+y=k$ $xy=l$ and this is equivalent $9k^2-18l+9+6l+4=2005$ <=> $3k^2=4(l+166)$ => $k=even$. The above is equivalent to $2k^2+k^2-4l-664=0$ (1) . The number $k^2-4l$ is the discriminant of $a^2-ka+l=0$ and to have roots it should be $\geq 0$. So $k^2-4l\geq 0$. So from (1) we have that $2k^2\leq 664$ <=> $k\leq 18,..$ so $k\leq 18$ because $k$ is an integer. Now if $k=18$ then from (1) we have that $l=77$ and $x=7$ or $11$ or $y=7$ or $11$. For $k=16$ we have $l=1044$ and discriminant$<0$ ,no roots If $k\leq 14$ , $3k^2<664$ and so $l<0$ absurb. So the only roots are $x=7$ or $11$,$y=7$ or $11$

The given equation is equivalent to $ 2(x+y)^2+(x-y)^2=664 $ Since $ x+y $ and $ x-y $ have the same parity they must be even. Now let $ x+y=2a $ and $ x-y=2b $. We have $ 16a^2+4b^2=664, 2a^2+b^2=166, b=2c, a^2+2c^2=83 $. Since $ a^2<83, a<9 $. Now it is easy to check that the solutions are $ (x,y)=(11,7) and (7,11) $ Another way of proving this problem is considering the equation as quadratic $ 3x^2+xy2+3y^2-664=0 $, so the discriminant must be perfect square, but doing it this way takes too much time.

Simplifying the equation we get $ 3x^2 + 3y^2 + 2xy = 664 $ or $3(x-y)^2 = 664-8xy = 8(83-xy)$ implying that $x$ and $y$ are odd. Letting $x=2a+1$ and $y = 2b+1$ we get $3(a-b)^2 = 2(83-xy) < 2\cdot 72$ or $(a-b)^2 < 46$ leaving $(a-b) = 6,4,2$ since $a$ and $b$ have the same parity. Now plugging these three cases in the equation we get $(x,y) \in \{(7,11),(11,7)\}$ .

How about AM-GM and common divisors: By AM-GM $2005 = 9(x^2+y^2+1) + 2(3xy+2) \ge 9(2xy+1) +2(3xy+2) =24xy+13 $ which gives $xy \le 83$ Through letting $x=ad;y=bd, (a,b)=1 $ we see that $d=1,2$ If $d=2$ then $166 = 3a^2 +3b^2 +8ab \ge 14 ab \rightarrow ab \le 11$ and checking cases no solution. Thus $d=1$. Now letting $y=x+a$ we see that $8x^2+8ax+3a^2=664$ which implies $a$ is a multiple of four. We now check coprime odd numbers that differ by a multiple of four and multiply to less than $83$, and we will obtain the result. Super-Ugly and bashy yet ...

We can re-write the equation as: $ 3x^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005$ or $ (3x + y)^2 = 4(498 - 2y^2)$ The above equation tells us that $(498 - 2y^2)$ is a perfect square. Since $498 - 2y^2 \ge 0$. this implies that $y \le 15$ Also, taking $mod 3$ on both sides we see that $y$ cannot be a multiple of $3$. Also, note that $249 - y^2$ has to be even since $(498 - 2y^2) = 2(249 - y^2)$ is a perfect square. So, $y^2$ cannot be even, implying that $y$ is odd. So we have only $\{1, 5, 7, 11, 13\}$ to consider for $y$. Trying above 5 values for $y$ we find that $y = 7, 11$ result in perfect squares. Thus, we have $2$ cases to check: $Case 1: y = 7$ $ (3x + 7)^2 = 4(498 - 2(7^2))$ $ => (3x + 7)^2 = 4(400)$ $ => x = 11$ $Case 2: y = 11$ $ (3x + 11)^2 = 4(498 - 2(11^2))$ $ => (3x + 11)^2 = 4(256)$ $ => x = 7$ Thus all solutions are $(7, 11)$ and $(11, 7)$.

Ok so a lil bit bashy but I bet it will be quick and won't be rigorous at all. Write the given equation as 3$(x-y)^{2}$ + 6$(x+y)^{2}$ = 1992 Now just put x-y as a and x+y as b. So bound like 1992>6$b^{2}$. and then check , there wont be that much You would be lucky if you start from the last like the max value b can take is 18, So you get the solns in this manner.

Solution expand and get $3x^2+3y^2+2xy=664$ Then make quad in $x$ Thus $7968-32y^2=k^2$ Thus $249-y^2=2a^2$ Now we know y is odd and less than $15$ so we can take $y=1,3,5,7,9,11,13,15 8$ cases And we get $y$ is $7$ or $11$ putting in original When $y =7$ Solution expand get $x =11$ we know qtn statement is symmetric thus WLOG if $y =11$ then $x=7$ done $(7,11)(11,7)$