If two polynomials both of n degree have n identical real roots, are they equal? Proof? 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?

Clarification:  To come to the conclusions in the answers we must assume the multiplicity of each root matches.  For example, It seems all the answers rule this possibility out.  For example, one can make a case saying these cubics have the same roots of 1 and -1.  However, they are obviously not the same polynomial.
$$f(x)=(x-1)(x-1)(x+1)$$ and
$$g(x)=(x-1)(x+1)(x+1)$$
Essentially, add the words "identical, including multiplicity, real roots" to rule out any possibility of confusion.
