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?

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


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

  • $\begingroup$ @robjohn: Thanks for your edit. $\endgroup$ – 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.


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.


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