17. Let $p(x)$ be a polynomial with integer coefficients such that $p(m)-p(n)$ divides $m^{2}-n^{2}$ for all integers $m$ and $n$. If $p(0)=1$ and $p(1)=2$, find the largest possible value of $p(100)$.
17. Answer: 10001.
Let $n=0$. Since $p(m)-1 \mid m^{2}$ for all $m, \operatorname{deg} p(x) \leq 2$. Let $p(x)=a x^{2}+b x+c$. Then
$$
\frac{p(m)-p(n)}{m-n}=\frac{a\left(m^{2}-n^{2}\right)+b(m-n)}{m-n}=a(m+n)+b
$$
divides $m+n$. If $a \neq 0$, then $a= \pm 1$ and $b=0$; if $a=0$, then $b= \pm 1$. Thus
$$
p(x)= \pm x^{2}+c, \pm x+c, c \text {. }
$$

Since $p(0)=1$ and $p(1)=2$, we have $p(x)=x^{2}+1$ or $p(x)=x+1$. The largest possible value of $p(100)$ is $100^{2}+1=10001$.