451 Find all integer solutions to the indeterminate equation
$$
\left(x^{2}-4\right)\left(y^{2}-4\right)=t^{2}+12
$$
If $x=0$, then
$t^{2}+4 y^{2}=4 \Rightarrow t=0, y= \pm 1$ or $t= \pm 2, y=0$.
If $y=0$, similarly,
$t=0, x= \pm 1$ or $t= \pm 2, x=0$.
If $t=0$, then $x= \pm 1, y=0$ or $x=0, y= \pm 1$.
Consider the non-zero integer solutions of the equation. It is easy to see that if the equation has non-zero integer solutions, then there must exist a set of positive integer solutions.
We only need to prove that the original equation has no positive integer solutions.
If the equation has a set of positive integer solutions $\left(x_{0}, y_{0}, t_{0}\right)$, then $x_{0}^{2} y_{0}^{2}-4\left(x_{0}^{2}+y_{0}^{2}-1\right)=t_{0}^{2}$.
This indicates that the indeterminate equation $x^{2}+y^{2}+z^{2}=x y z+1$ has positive integer solutions.

Let $00$, i.e., $\left(x_{0}, y_{0}, z_{1}\right)$ is also a positive integer solution of the equation.
By the minimality condition, we know $z_{1}=x_{0} y_{0}-z_{0} \geqslant z_{0}$, i.e.,
$$
\frac{z_{0}}{x_{0} y_{0}} \leqslant \frac{1}{2} \text{. }
$$

If $x_{0}=1$, then $y_{0}^{2}+z_{0}^{2}=y_{0} z_{0}$ has no solution;
If $x_{0}=2$, then $y_{0}^{2}+z_{0}^{2}+4=2 y_{0} z_{0}+1$, i.e., $\left(y_{0}-z_{0}\right)^{2}=-3$ also has no solution.
Thus, $x_{0} \geqslant 3$.
Therefore, $z_{0}=\frac{x_{0}}{y_{0}}+\frac{y_{0}}{x_{0}}+\frac{z_{0}^{2}}{x_{0} y_{0}}-\frac{1}{x_{0} y_{0}}<1+\frac{z_{0}}{3}+\frac{z_{0}}{2}$ $\Rightarrow z_{0}<6$.
Substituting $z_{0}=3,4,5$ into the original equation yields no solutions, hence, the original equation has no positive integer solutions.
In summary, all integer solutions of the original equation are
$$
(x, y, t)=(0, \pm 1,0),(0,0, \pm 2),( \pm 1,0,0) .
$$