# Simple polynomial proof

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

Problem 165. Assume that

$$\left\{\begin{array}116a+4b+c=0\\49a+7b+c=0\\100a+10b+c=0\end{array}\right.$$

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:

$$\left\{\begin{array}(x_1^2a+x_1b+c=0\\x_2^2a+x_2b+c=0\\x_3^2a+x_3b+c=0\end{array}\right.$$

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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

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.

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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 $x_1\neq x_2 \neq x_3 \neq x_1$.

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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