One, (20 points) Given the quadratic equation $x^{2}+a x+b=0$ has two consecutive integer roots, and the quadratic equation $x^{2}+b x+a=0$ has integer roots. Find the values of $a, b$.


Let the two roots of $x^{2}+a x+b=0$ be $n$ and $n+1$ (where $n$ is an integer). Then
$$
2 n+1=-a, \quad n(n+1)=b.
$$

Thus, the equation $x^{2}+b x+a=0$ can be rewritten as
$$
x^{2}+n(n+1) x-(2 n+1)=0.
$$
(1) If $n \geqslant 0$, then equation (1) has one positive and one negative root. Let the positive integer root be $m$, then
$$
m^{2}+n(n+1) m-(2 n+1)=0,
$$

which simplifies to $m n^{2}+(m-2) n+\left(m^{2}-1\right)=0$.
Since $n$ is an integer, we have
$$
\Delta=(m-2)^{2}-4 m\left(m^{2}-1\right) \geqslant 0.
$$

Rearranging gives $4 m \leqslant \frac{4}{m^{2}}+1$.
If $m \geqslant 2$, the above inequality does not hold. Hence, $0<m<2$.
Thus, $m=1$.
Substituting $m=1$ into equation (2) yields $n^{2}-n=0$.
Solving this, we get $n=0,1$.
When $n=0$, $a=-1, b=0$;
When $n=1$, $a=-3, b=2$.
(2) If $n \leqslant-1$, then equation (1) has two negative integer roots, i.e.,
$$
\Delta=n^{2}(n+1)^{2}+4(2 n+1)
$$

must be a perfect square and even. Since $n \leqslant-1$, then
$$
n^{2}(n+1)^{2}+4(2 n+1)<[n(n+1)]^{2}.
$$

Thus, $n^{2}(n+1)^{2}+4(2 n+1) \leqslant[n(n+1)-2]^{2}$.
Therefore, $n^{2}+3 n \leqslant 0$, i.e., $n=-3,-2,-1$.
Upon verification, when $n=-1,-2$, equation (1) has no real roots;
When $n=-3$, equation (1) becomes $x^{2}+6 x+5=0$.
Its roots are $-1,-5$, at this time, $a=5, b=6$.
Combining (1) and (2), we have
$$
(a, b) \text{ are } (-1,0), (-3,2), (5,6).
$$