Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

this problem is from I.M Gelfand's book called Algebra.

Problem 165. Assume that


Prove that $a=b=c=0$

I know I could simply solve this equation for $a,b,c$ and show that they're equal to 0 but I wonder if there's more smart way of showing it, since this system looks like:


share|cite|improve this question
Try Vandermonde determinant. – copper.hat Nov 8 '12 at 17:18
Should the first equation read $16a+\cdots$ instead of $6a+\cdots$? – Hagen von Eitzen Nov 8 '12 at 17:20
Yes Hagen, sorry, I will correct that now. I don't know about Vandermonde matrix, but I will read about it now. I hope I will understand. – Paul Dirac Nov 8 '12 at 17:26
Yes, your "wondering" is on the right track, since that means it's a corollary to your prior question, on the prior Problem 164. – Bill Dubuque Nov 8 '12 at 17:43
up vote 2 down vote accepted

Once you view it that way, you can quickly see that if you have a solution $(a,b,c)$ then the equation $ax^2+bx+c=0$ would have three solutions, $4$,$7$ and $10$. But if $(a,b,c)\neq(0,0,0)$ then $ax^2+bx+c$ is a non-zero polynomial of degree smaller than $3$, which cannot have $3$ distinct roots.

Hopefully, you can see that it works for any number of variables. If you have $n$ distinct values $x_1,x_2,...,x_n$ there is no non-zero sequence of values $(a_0,..,a_{n-1})$ such that for $i=1,...n$, $\sum_{j=0}^{n-1} a_j x_i^j = 0$.

If there were such a sequence, then $p(x)=\sum_{j=0}^{n-1} a_ix^i$ would be a non-zero polynomial with $\deg(p(x))<n$ but with $n$ distinct roots.

share|cite|improve this answer
So that proves that $a,b,c$ has to be equal to 0 right? – Paul Dirac Nov 8 '12 at 17:36
Yes, by contradiction. – Thomas Andrews Nov 8 '12 at 17:37
Thank you for help Thomas Andrews, I will add you reputation once I reach 15 reps. – Paul Dirac Nov 8 '12 at 17:39
This method is more than cool! – Hagen von Eitzen Nov 8 '12 at 17:45

From this the system

$$ \begin{array}(x_1^2a+x_1b+c=b_1 \\ x_2^2a+x_2b+c=b_2\\ x_3^2a+x_3b+c=b_3 \end{array} $$

has a unique solution (in $a,b,c$) iff $x_1\neq x_2 \neq x_3 \neq x_1$.

share|cite|improve this answer
Thank you for help but I have very basic knowledge on matrices in general so I don't think I will understand that proof. – Paul Dirac Nov 8 '12 at 17:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.