11. (20 points) The quadratic function $y=f(x)$ has a coefficient of 1 for the quadratic term, and both the linear term 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. Try to find the function $f(x)$, such that the sum of its coefficients is minimized.


11. Let $f(x)=(x-a)^{2}-b^{2}(b>0)$, then the four roots of $f(f(x))=0$ are $a \pm \sqrt{a+b+b^{2}}$, $a \pm \sqrt{a-b+b^{2}}$. Therefore, according to the problem, we have
$$
3 \sqrt{a-b+b^{2}}=\sqrt{a+b+b^{2}},
$$

which means
$$
4 a=5 b-4 b^{2} .
$$

Also, $2 a \in \mathbf{Z}, a^{2}-b^{2} \in \mathbf{Z}$, so $4 a^{2}-4 b^{2} \in \mathbf{Z}, 4 a^{2} \in \mathbf{Z}$, hence $4 b^{2} \in \mathbf{Z}$.
From equation (1) and $2 a \in \mathbf{Z}, 4 b^{2} \in \mathbf{Z}$, we get $5 b \in \mathbf{Z}$, then $25 b^{2} \in \mathbf{Z}$, and thus $25 b^{2}-6 \times 4 b^{2}=b^{2} \in \mathbf{Z}$.
From $5 b \in \mathbf{Z}, b^{2} \in \mathbf{Z}$, we know $b \in \mathbf{Z}$.
From $a^{2}-b^{2} \in \mathbf{Z}, b^{2} \in \mathbf{Z}$, we get $a^{2} \in \mathbf{Z}$, and since $2 a \in \mathbf{Z}$, we have $a \in \mathbf{Z}$.
From $a=\frac{5 b}{4}-b^{2} \in \mathbf{Z}$, we know $4 \mid k$, so $b=4 k\left(k \in \mathbf{N}^{*}\right), a=5 k-16 k^{2}$. Therefore,
$$
f(x)=x^{2}+2\left(16 k^{2}-5 k\right) x+\left(16 k^{2}-5 k\right)^{2}-16 k^{2},
$$

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

Therefore, when $k=1$, the sum of the coefficients of $f(x)$ is minimized. At this time, $f(x)=x^{2}+22 x+105, f(f(x))$ has four roots: $-8,-10,-12,-14$.