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 $P(x) - Q(x) = 0 \implies P(x) = Q(x)$.
