Problem 169 from the book I.M. Gelfand, "Algebra".

"Assume that $x_1, \ldots , x_{10}$ are different numbers, and $y_1 , \ldots , y_{10}$ are arbitrary numbers. Prove that there is one and only one polynomial $P(x)$ of degree not exceeding $9$ such that $P(x_1) = y_1,$ $P(x_2) = y_2,\ldots, P(x_{10}) = y_{10}$."

I'm dubious about my proof of uniqueness:

I will prove it by contradiction, I will suppose that there are at least two such polynomials of the same degree and will show that they're the same. Suppose that there're more than one such polynomial of degree 9, denote them as $P(x)$ and $Q(x)$. So I can get a new polynomial $Z(x) = P(x) - Q(x)$, which is either zero or a new polynomial of degree not exceeding 9. Polynomial $Z(x)$ for values $x1, ..., x10$ has to yield $0$, because it is a subtraction of two polynomials that for these values yield the same numbers. It means that $Z(x)$ has 10 roots, but a 9-degree polynomial can have maximum 9 roots, so it is a zero polynomial, $Z(x) = 0$. In this case $P(x) - Q(x) = 0 \implies P(x) = Q(x)$.

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    $\begingroup$ Argument is fine. A polynomial of degree $\le 9$ has at most 9$ roots, unless it is the identically $0$ polynomial. For existence, use the Lagrange interpolating polynomial. $\endgroup$ – André Nicolas Jan 14 '13 at 7:20
  • $\begingroup$ These are polynomials over $\mathbb{R}$? You should be conscientious that you've used the fact that a polynomial of degree $n\geq 0$ has at most $n$ roots, and how to prove it. $\endgroup$ – user7530 Jan 14 '13 at 7:30
  • $\begingroup$ Yes, over real numbers. "the fact that a polynomial of degree n≥0 has at most n roots, and how to prove it." I just assume it, I'm more interested if I assume this fact, is my proof legitimate. $\endgroup$ – Graduate Jan 14 '13 at 8:04
  • $\begingroup$ Yes. Of course if you assume that there is a unique polynomial interpolating of degree $\leq n$ interpolating $n$ points, it would also be legitimate. $\endgroup$ – user7530 Jan 14 '13 at 17:24

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