5. If two integers $x, y$ satisfy the equation
$$
(2 x+9 y)^{2006}+(4 x-y)^{2006}=7777^{m} \text {, (1) }
$$

then the array $(x, y)$ is called an integer solution of equation (1). The number of integer solutions of equation (1) is ( ).
(A) 0
(B) 1
(C) 2
(D) 2006
5.A.

Assume there exist integers $x, y$ satisfying equation (1).
If $7 \mid (2x + 9y)$, then from equation (1) we know $7 \mid (4x - y)$. At this point, the left side of equation (1) can be divided by $7^{200}$, but the right side of equation (1) can only be divided by $7^m$.
Therefore, $7 \nmid (2x + 9y)$, $7 \nmid (4x - y)$.
For $\forall a \in \mathbf{Z}$,
$7 \mid [(a-3)(a-2)(a-1)a(a+1)(a+2)(a+3)]$
$\Rightarrow 7 \mid (a^7 - a) = a(a^6 - 1)$.
Thus, $7 \mid [(2x + 9y)^6 - 1]$, $7 \mid [(4x - y)^6 - 1]$.
Therefore, $7 \mid \{(2x + 9y)^2[(2x + 9y)^{6 \times 34} - 1]\}$,
$7 \mid \{(4x - y)^2[(4x - y)^{6 \times 334} - 1]\}$.
From equation (1) we know
$7 \mid [(2x + 9y)^2 + (4x - y)^2] \Rightarrow 7 \mid (x^2 + 2y^2)$.
It is easy to see that the remainder of any perfect square when divided by 7 is $0, 1, 2, 4$.
Upon inspection, only when $7 \mid x$, $7 \mid y$, does $7 \mid (x^2 + 2y^2)$. But in this case, $7 \mid (2x + 9y)$, $7 \mid (4x - y)$, which is a contradiction.
Therefore, there do not exist integers $x, y$ that satisfy equation (1).