mathisfun7 wrote: Let $k$ and $n$ be positive integers and let $x_1, x_2, \cdots, x_k, y_1, y_2, \cdots, y_n$ be distinct integers. A polynomial $P$ with integer coefficients satisfies \[P(x_1)=P(x_2)= \cdots = P(x_k)=54\] \[P(y_1)=P(y_2)= \cdots = P(y_n)=2013.\] Determine the maximal value of $kn$. So $P(x)=54+Q(x)\prod_i(x-x_i)$ and $1959=Q(y_j)\prod_i (y_j-x_i)$ Since $1959=3\times 653$ can be at most the product of $4$ distinct integers, we get $k,n\le 4$ A solution with $k=3$ and $n=2$ exists : $P(x)=54-653x(x-2)^2(x-4)$ with $\{x_i\}=\{0,2,4\}$ and $\{y_j\}=\{1,3\}$ If WLOG $k=4$ and $n\ge 2$, then : WLOG consider $y_1<y_2$ $y_1-x_1\in T=\{t_i\}=\{-1959,-653,-3,-1,1,3,653,1959\}$ and so $x_1=y_1-t_{i_1}\}$ $y_2-x_1\in T$ and so $x_1=y_2-t_{j_1}$ So $y_2-y_1=t_{j_1}-t_{i_1}>0$ Same, $x_2=y_1-t_{i_2}\}$ and $x_2=y_2-t_{j_2}$ and $y_2-y_1=t_{j_2}-t_{i_2}$ Same, $x_3=y_1-t_{i_3}\}$ and $x_3=y_2-t_{j_3}$ and $y_2-y_1=t_{j_3}-t_{i_3}$ Same, $x_4=y_1-t_{i_4}\}$ and $x_4=y_2-t_{j_4}$ and $y_2-y_1=t_{j_4}-t_{i_4}$ So $t_{j_1}-t_{i_1}=t_{j_2}-t_{i_2}=t_{j_3}-t_{i_3}=t_{j_4}-t_{i_4}>0$ Looking at $t_i$, no such pairs exist. The only posibilities are made of three pairs and are : $3-1=1-(-1)=(-1)-(-3)=2$ $1959-653=653-(-653)=(-653)-(-1959)=1306$ So $n\ge 2$ $\implies$ $k\le 3$ It remains to check the case $n=k=3$ But then : $x_1=y_1-t_{i_1}\}$ and $x_1=y_2-t_{j_1}$ and $y_2-y_1=t_{j_1}-t_{i_1}$ $x_2=y_1-t_{i_2}\}$ and $x_2=y_2-t_{j_2}$ and $y_2-y_1=t_{j_2}-t_{i_2}$ $x_3=y_1-t_{i_3}\}$ and $x_3=y_2-t_{j_3}$ and $y_2-y_1=t_{j_3}-t_{i_3}$ So $t_{j_1}-t_{i_1}=t_{j_2}-t_{i_2}=t_{j_3}-t_{i_3}>0$ And so $y_2-y_1\in\{2,1306\}$ But also $y_3-y_1\in\{2,1306\}$ and $y_3-y_2\in\{2,1306\}$ Impossible. Hence the answer : $\boxed{(kn)_{\text{MAX}}=6}$

p(a)-p(b) :a-b