11. (40 points) For the quadratic function $p(x)=a x^{2}+b x+c$, for some integers $n$, the equality $p(n)=p\left(n^{2}\right)$ holds. Provide an example of the function $p(x)$ for which the number of such numbers $n$ is the largest. What is this largest number of numbers $n$?
Answer: The maximum number of numbers $n$ is 4. An example of a function: $p(x)=x^{2}-6 x+1$.

Solution: Since $0=0^{2}$ and $1=1^{2}$, for any function $p(x)$ there are at least two such numbers $n$.

Write the equality $p(n)=p\left(n^{2}\right)$ and transform it:

$$
a n^{2}+b n+c=a n^{4}+b n^{2}+c \Leftrightarrow a n^{2}\left(n^{2}-1\right)=b n(1-n)
$$

Further, assuming that $n \neq 0$ and $n \neq 1$, we get that $b=-a n(n+1)$.

For each function $p(x)$, the value $-\frac{b}{a}$ is known and fixed. Therefore, the values of $n$ of interest can be obtained as solutions to the quadratic equation $n^{2}+n+\frac{b}{a}=0$. Its roots are $n_{1,2}=\frac{-1 \pm \sqrt{D}}{2}$, where $D=1-4 \frac{b}{a}$. It is not difficult to see that the roots of the equation will be integers only if the numerator of the fraction is even, which means the discriminant must be a perfect square of some odd number; and in this case, we will have two integer solutions to the equation; in all other cases, there will be no integer roots.

The relationship between $n_{1}$ and $n_{2}$ follows from Vieta's theorem: $n_{1}+n_{2}=-1$, i.e., $n_{2}=-1-n_{1}$.

Thus, the set of numbers for which the condition of the problem is satisfied is the set $\left\{0,1, n_{1},-1-n_{1}\right\}$. Now let's determine whether there are such $n_{1}$ that this set consists of four numbers. It is easy to see that the conditions

$$
\left\{\begin{array} { l } 
{ n _ { 1 } \neq 0 , } \\
{ n _ { 1 } \neq 1 , } \\
{ - 1 - n _ { 1 } \neq 0 , } \\
{ - 1 - n _ { 1 } \neq 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
n_{1} \neq 0 \\
n_{1} \neq 1 \\
n_{1} \neq-1 \\
n_{1} \neq-2
\end{array}\right.\right.
$$

Obviously, such $n_{1}$ exist, and therefore, the maximum number of numbers $n$ for which the condition of the problem is satisfied is 4. To construct an example, let $n_{1}=2, a=1$. We get that $n_{2}=-1-n_{1}=-3$ and $b=-a n_{1}\left(n_{1}+1\right)=-1 \cdot 2 \cdot 3=-6$. Let $c=1$. Then we get that $p(x)=x^{2}-6 x+1$. Let's check: $p(0)=p\left(0^{2}\right)=1$, $p(1)=p\left(1^{2}\right)=-4, p(2)=p\left(2^{2}\right)=-7, p(-3)=p\left((-3)^{2}\right)=28$.