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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?

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They may differ by a constant factor. For example, $2x^2-2\neq x^2-1$, but they have identical roots. –  Mårten W Dec 26 '13 at 8:52
    
That's an important fact I forgot to add - doing it now.\ –  user116791 Dec 26 '13 at 9:04

2 Answers 2

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 $A\not=0\in\mathbb R$.

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@robjohn: Thanks for your edit. –  mathlove Dec 26 '13 at 9:04

OP here; I proved it succesfully.

By the Polynomial Remainder theorem (http://en.wikipedia.org/wiki/Polynomial_remainder_theorem), if any n-degree polynomial P has n roots, then P must be divisible by (x-r) where r is each of the n roots.

Thus P = k * (x- r1)(x - r2)...(x - rn) where k is some value.

Now since P is of n degree and there are n roots, k must be of degree 0, otherwise P's degree would be greater than k. So P = k * (x- r1)(x - r2)...(x - rn) where k is some integer.

Any two polynomials P1 and P2 of identical degree of roots can be represented as k1 * (x- r1)(x - r2)...(x - rn) and k2 * (x- r1)(x - r2)...(x - rn) ... since k1 and k2 are both integers, the two polynomials must be some linear multiple of each other. So the proof is complete.

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