Three. (25 points) Find real numbers $a, b$, such that the function $y_1 = x^2 + ax + b$, $y_2 = x^2 + bx + a$ and the x-axis have four intersection points, where the distances between adjacent points are equal.

---

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Three, from the problem, we know $a \neq b$. Without loss of generality, let $a > b$, and set $y_{1}=y_{2}$, then $x^{2}+a x+b=x^{2}+b x+u$.
We get $x=1$, so $y_{1}$ and $y_{2}$ intersect only at the point $(1, a+b+1)$.
The graphs of $y_{1}$ and $y_{2}$ that meet the problem's conditions have 4 forms:
Figure 5

Without loss of generality, let $y_{1}$ intersect the $x$-axis at $\mathrm{A}\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $y_{2}$ intersect the $x$-axis at $\left(x_{3}, 0\right)$ and $\left(x_{4}, 0\right)$. Then
$$
\begin{array}{l}
\Delta_{1}=a^{2}-4 b, \Delta_{2}=b^{2}-4 a . \\
x_{1,2}=\frac{-a \pm \sqrt{\Delta_{1}}}{2}, x_{3,4}=\frac{-b \pm \sqrt{\Delta_{2}}}{2} .
\end{array}
$$

For Figure 5(1), by the problem's conditions, $|A C|=|B I|$, we have
$$
x_{1}-x_{3}=x_{4}-x_{2}, x_{1}+x_{2}=x_{3}+x_{4} \text {, }
$$

Then $u=b$ (which does not meet the problem's conditions).
For Figure 5(2), it is the same.
For Figures 5(3) and 5(4), we always have $|A B|=|I|$, i.e.,
$$
\sqrt{\Delta_{1}}=\sqrt{\Delta_{2}} \Rightarrow \sqrt{u^{2}-4 b}=\sqrt{b^{2}-4 u} \text {. }
$$

We get $a^{2}-b^{2}+4 a-4 b=0$, then $a=b$ (which does not meet the problem's conditions), $u+b=-4$. Therefore, $v_{1}$ and $v_{2}$ intersect at the fixed point $(1,-3)$. Thus, Figure 5(4) is impossible.
For Figure 5(3), we have $|A C|=|B C|$, i.e.,
$$
\begin{array}{l}
x_{3}-x_{1}=x_{2}-x_{3}, 2 x_{3}=x_{1}+x_{2}, \\
-b-\sqrt{\Delta_{2}}=-u, \\
\text { Hence } \sqrt{b^{2}-4 u}=u-b .
\end{array}
$$

By $a+b=-4$, we know $\sqrt{(a+4)^{2}-4 a}=2 a+4$. Squaring and rearranging, we get $u^{2}+4 a=0$, so $\left\{\begin{array}{l}a=0, \\ b=-4 .\end{array}\right.$
Upon verification, this solution meets the problem's conditions.
By symmetry, it is easy to know $\left\{\begin{array}{l}a=-4 \\ b=0\end{array}\right.$. This also meets the problem's conditions.
(Proposed by Xu Changhai, Xiangtai Vocational School, Lai'an County, Anhui Province)