14. (2004 Australian Mathematical Olympiad) Find all integer pairs $(a, b)$ such that $\left(a^{3}+b\right)\left(a+b^{3}\right)=(a+b)^{4}$.
14. Notice that
$$
\begin{array}{l}
\left(a^{3}+b\right)\left(a+b^{3}\right)=(a+b)^{4} \\
\Leftrightarrow a^{4}+a^{3} b^{3}+a b+b^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\
\Leftrightarrow a^{3} b^{3}+2 a^{2} b^{2}+a b=4 a^{3} b+8 a^{2} b^{2}+4 a b^{3} \\
\Leftrightarrow a b(a b+1)^{2}=4 a b(a+b)^{2} \\
\Leftrightarrow a b\left[(a b+1)^{2}-4(a+b)^{2}\right]=0
\end{array}
$$

is true. Therefore, $(a, 0)$ and $(0, b)$ are solutions to the given equation, where $a, b \in \mathbf{Z}$.
Other solutions must satisfy $(a b+1)^{2}-4(a+b)^{2}=0$.
Since $(a b+1)^{2}-4(a+b)^{2}=0$, we have
$$
a b+1= \pm 2(a+b) \text {. }
$$

We consider two cases.
If $a b+1=2(a+b)$, then
$$
(a-2)(b-2)=3 \text {. }
$$

Thus, we have
$$
\left\{\begin{array} { l } 
{ a - 2 = 3 , } \\
{ b - 2 = 1 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ a - 2 = 1 , } \\
{ b - 2 = 3 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ a - 2 = - 3 , } \\
{ b - 2 = - 1 }
\end{array} \text { or } \left\{\begin{array}{l}
a-2=-1, \\
b-2=-3 .
\end{array}\right.\right.\right.\right.
$$

Solving these, we get
$$
a=5, b=3 ; a=3, b=5 ; a=-1, b=1 ; a=1, b=-1 \text {. }
$$

If $a b+1=-2(a+b)$, then
$$
(a+2)(b+2)=3 \text {. }
$$

Similarly, we get
$$
a=1, b=-1 ; a=-1, b=1 ; a=-5, b=-3 ; a=-3, b=-5 \text {. }
$$

In summary, the set of all possible solutions to the given equation is
$$
\{(a, 0) \mid a \in \mathbf{Z}\} \cup\{(0, b) \mid b \in \mathbf{Z}\} \cup\{(-5,-3),(-3,-5),(-1,1),(1,-1),(3,5),(5,3)\}
$$