CORRECTION: The polynomials don't have to be equal, but one has to be a constant multiple of the other.
I ask the question because I saw this fact used in this solution to a problem:
Problem:
Given that
$ \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 $
$ \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 $
$ \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 $
$ \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 $
find the value of $ w^2+x^2+y^2+z^2 $
Here's the solution
http://www.isinj.com/aime/AIME-Solutions-1983-2011.pdf
Starts at the bottom of page 22, problem #15.
On page 23, the solution compares the LHS of eq (2) to the LHS of eq (3) because they are both 4th degree polynomials with 4 identical roots. I am trying to prove that this must be true for all polynomials. Can someone help me prove it?
Best Answer
HINT :
If a polynomial has $n$ roots $\alpha_i\ (i=1,2,\cdots,n)$, it is represented as $$f(x)=A(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_n)$$ where $A\not=0\in\mathbb R$.