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ThePerfectHacker · #6 December 11th, 2007, 10:37 AM

I came up with problem #1 accidently when I playing around with polynomials. Here is my original solution, however it seems to me that this problem is easy even if approached directly.

Proof:
Let $a_1,...,a_n$ be real (or complex numbers) and define $f(x) = (x-a_1)...(x-a_n)$ . The key step is to note that $f'(x_k) = P_k$ by using the general product rule for derivatives. Now, $\deg f(x) = n\geq 2$ thus, $\deg f'(x) = n-1\geq 1$ , this means that $f'(x)$ cannot attain the same values at $a_1,a_2,...,a_n$ (meaning $f(a_1)=f(a_2)=...=f(a_n)$ ) because otherwise the situation is that a degree $n-1$ polynomial attains the same value $n$ times for $n$ distinct numbers. Which is impossible. Thus, $P_1,P_2,...,P_n$ cannot all be the same.