Let no three of $x_{1},x_{2},\cdots,x_{1996}$ are same. $w_{n}(x)=x^2-nx$ then there exist $ x_{i}, x_{j}, x_{k}$ s.t. $w_{n}(x_{i})=w_{n}(x_{j})=w_{n}(x_{k})$ WLOG assume $x_{i}\neq x_{j}$ then $x_{i}+x_{j}=n$ so, for given any n, there is $x_{i},x_{j}$ s.t. $x_{i}+x_{j}=n$ but $x_{i}+x_{j}$ can be atmost $\binom{1996}{2}$ distinct values. so, contradiction. so, what we assume is wrong. so, atleast three of $x_{1},x_{2},\cdots,x_{1996}$ are same.

sumanguha wrote: then $x_{i}+x_{j}=n$ why $x_{i}+x_{j}=n$ ? we do not know $x_{i} , x_{j}$ are the root of $ w_{n}(x)$ ! ________________________________________________________________________ here is my solution : assume ${w_{x_{i}}(x)= (x-x_{i}})^2$ we know $w_{x_{1}}(x_{i})=w_{x_{1}}(x_{j})=w_{x_{1}}(x_{k})$ if $w_{x_{1}}(x_{i})=0$ it means that $x_{i}=x_{j}=x_{k}=x_{1}$ and we are done . if $w_{x_{1}}(x_{i})\neq 0$ it means that $ x_{1}-x_{k}=x_{j}-x_{1}$ now put $w_{x_{k}}(x)$ and do this again . we can do this intolerable and after one we get that exist a new $x_{i}$ so, contradiction .

Quote: why $x_{i}+x_{j}=n$ ? we do not know $x_{i}$,$x_{j}$ are the root of $w_{n}(x)$ ! because we assume no three of $x_{1}, x_{2},..., x_{1996}$ are same so, atleast two of $x_{i}, x_{j}, x_{k}$ are different for $w_{n}(x_{i})= w_{n}(x_{j}) =w_{n}(x_{k}) $ wlog assume $x_{i}\neq x_{j}$ so, $x_{i}+x_{j}=n$

Let $m=min(x_{1},\dots ,x_{1996}), W(x)=(x-m)^2$ and $W(x_i)=W(x_j)=W(x_k)$.Then $x_i=x_j=x_k$.The proof is completed.$\blacksquare$