Solution 1
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 $(11, 7)$ .
$Kris17$
Solution 2
Expanding, combining terms, and factoring results in $$\begin{align*} 9x^2 + 9y^2 + 9 + 6xy + 4 &= 2005 \\ 9x^2 + 9y^2 + 6xy &= 1992 \\ 3x^2 + 3y^2 + 2xy &= 664 \\ (x+y)^2 + 2x^2 + 2y^2 &= 664. \end{align*}$$ Since $2x^2$ and $2y^2$ are even, $(x+y)^2$ must also be even, so $x$ and $y$ must have the same parity. There are two possible cases.
Case 1: and are both even
Let $x = 2a$ and $y = 2b$ . Substitution results in $$\begin{align*} 4(a+b)^2 + 8a^2 + 8b^2 &= 664 \\ (a+b)^2 + 2a^2 + 2b^2 &= 166 \end{align*}$$ Like before, $a+b$ must be even for the equation to be satisfied. However, if $a+b$ is even, then $(a+b)^2$ is a multiple of 4. If $a$ and $b$ are both even, then $2a^2 + 2b^2$ is a multiple of 4, but if $a$ and $b$ are both odd, the $2a^2 + 2b^2$ is also a multiple of 4. However, $166$ is not a multiple of 4, so there are no solutions in this case.
Case 2: and are both odd
Let $x = 2a+1$ and $y = 2b+1$ , where $a,b \ge 0$ . Substitution and rearrangement results in $$\begin{align*} 4(a+b+1)^2 + 2(2a+1)^2 + 2(2b+1)^2 &= 664 \\ 2(a+b+1)^2 + (2a+1)^2 + (2b+1)^2 &= 332 \\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \end{align*}$$ Note that $3a^2 \le 164$ , so $a \le 7$ . There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$ . Since both $a$ and $b$ are integers, we must have $$\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\ &= -32a^2 - 32a + 16 + 12 \cdot 164 \\ &= 16(-2a^2 - 2a + 1 + 3 \cdot 41) \\ &= 16(-2(a^2 + a) + 124), \end{align*}$$ where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square.
After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$ . If $a = 3$ , then $b = 5$ , and if $a = 5$ , then $b = 3$ .
Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\boxed{(7,11) , (11,7)}$ .