For the quadratic function $y=f(x)$ with a leading coefficient of 1, and both the linear coefficient and the constant term are integers. If $f(f(x))=0$ has four distinct real roots, and these roots can be arranged in an arithmetic sequence, find $f(x)$ such that the sum of its coefficients is minimized.

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Let $f(x)=(x-a)^{2}-b^{2}(b>0)$.
Then the four roots of $f(f(x))=0$ are
$$
a \pm \sqrt{a+b+b^{2}}, a \pm \sqrt{a-b+b^{2}}.
$$

From the analysis, we get
$$
3 \sqrt{a-b+b^{2}}=\sqrt{a+b+b^{2}},
$$

which implies $4 a=5 b-4 b^{2}$.
Given $2 a \in \mathbf{Z}$ and $a^{2}-b^{2} \in \mathbf{Z}$,
$\Rightarrow 4 a^{2}-4 b^{2} \in \mathbf{Z}, 4 a^{2} \in \mathbf{Z}$
$\Rightarrow 4 b^{2} \in \mathbf{Z}$.
From equations (1) and (2), and $2 a \in \mathbf{Z} \Rightarrow 5 b \in \mathbf{Z}$.
Also, from equation (2), $b \in \mathbf{Z}$.
Furthermore, $a^{2}-b^{2} \in \mathbf{Z} \Rightarrow a^{2} \in \mathbf{Z}$.
Since $2 a \in \mathbf{Z} \Rightarrow a \in \mathbf{Z}$.
Noting that, $a=\frac{5 b}{4}-b^{2} \in \mathbf{Z}$.
Thus, $b=4 k\left(k \in \mathbf{Z}_{+}\right), a=5 k-16 k^{2}$,
$$
\begin{array}{l}
f(x)=x^{2}+2\left(16 k^{2}-5 k\right) x+\left(16 k^{2}-5 k\right)^{2}-16 k^{2} \\
=0,
\end{array}
$$

The sum of the coefficients is
$$
\begin{aligned}
& 1+2\left(16 k^{2}-5 k\right)+\left(16 k^{2}-5 k\right)^{2}-16 k^{2} \\
= & \left(16 k^{2}-5 k\right)^{2}+16 k^{2}-10 k+1 .
\end{aligned}
$$

When $k=1$, the sum of the coefficients is minimized,
$$
f(x)=x^{2}+22 x+105.
$$

Upon verification, the four roots are $-8, -10, -12, -14$.