12. Satisfy:
$$
x+y^{2}=z^{3}, x^{2}+y^{3}=z^{4}, x^{3}+y^{4}=z^{5}
$$

the number of ordered triples of real numbers $(x, y, z)$ is $\qquad$.
12.7.

From the problem, we have
$$
\begin{array}{l}
\left(x+y^{2}\right)\left(x^{3}+y^{4}\right)=z^{8}=\left(x^{2}+y^{3}\right)^{2} \\
\Rightarrow x y^{2}(x-y)^{2}=0 .
\end{array}
$$

If $x=0$, then $y^{2}=z^{3}, y^{3}=z^{4}$. In this case, the only solutions are $(x, y, z)=(0,0,0)$ or $(0,1,1)$.
If $x \neq 0, y=0$, then $x=z^{3}, x^{2}=z^{4}$. In this case, the only solutions are $(x, y, z)=(1,0,1)$ or $(-1,0,-1)$.
If $x=y \neq 0$, then $x+x^{2}=z^{3}, x^{2}+x^{3}=z^{4}$.
If $x=y=-1$, then $z=0$. Otherwise,
$$
z=\frac{x^{2}+x^{3}}{x+x^{2}}=x,
$$

In this case, from $x+x^{2}=x^{3}$, we get
$$
x=y=z=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$

Therefore, there are 7 such ordered triples of real numbers.