5. The number of integer solutions $(x, y)$ for the equation $x^{3}+6 x^{2}+5 x=y^{3}-y+2$ is ( ).
(A) 0
(B) 1
(C) 3
(D) infinitely many
5.A.

The original equation can be rewritten as
$$
\begin{array}{l}
x(x+1)(x+2)+3\left(x^{2}+x\right) \\
=y(y-1)(y+1)+2 .
\end{array}
$$

Since the product of three consecutive integers is a multiple of 3, the left side of the equation is a multiple of 3, while the right side leaves a remainder of 2 when divided by 3, which is impossible. Therefore, the original equation has no integer solutions.