# Unique polynomial of degree at most 4 with rational coefficients

I'm working on the following problem:

Prove that there is a unique polynomial $$f$$ of degree at most $$4$$ with rational coefficients such that $$f(1) = 1, f(2) = 2, f(3) = 4, f(4) = 8,$$ and $$f(5) = 16$$.

I have a proof of a more general fact that there is a unique polynomial of degree at most $$n$$ that passes through $$n+1$$ data points:

Suppose that there are at least two polynomials of degree at most $$n$$ that pass through the $$n + 1$$ data points $$(x_0, y_0), ... , (x_n, y_n)$$, $$P(x)$$ and $$Q(x)$$. Define $$R(x) = P(x) - Q(x)$$. Since both $$P(x)$$ and $$Q(x)$$ pass through the $$n + 1$$ data points, $$P(x_i) = Q(x_i)$$ ($$i = 0,...,n$$) $$\Rightarrow$$ $$R(x_i) = P(x_i) - Q(x_i) = 0$$ ($$i = 0,...,n$$). $$R(x)$$ is an $$n$$th order polynomial with $$n + 1$$ zeroes $$\Rightarrow$$ $$R(x)$$ is the zero polynomial $$\Rightarrow$$ $$P(x) = Q(x)$$.

Can I simply rehash this proof for the above problem, simply by having both $$P(x)$$ and $$Q(x)$$ be polynomials with rational coefficients and of degree at most $$n = 4$$ ? Or is there an entirely different proof required here? If so, how can I go about proving the above problem?

Thanks!

• No, you can't, that would be a circular argument (unless the theorem is constructible, i.e. it gives you an algorithm to construct a polynomial passing through the data points). You would start with assuming that such $P$ and $Q$ exist, only to prove that such $P$ and $Q$ exist and $P=Q$. After you exhibit such a polynomial $P$, you can use the theorem to show that it is unique. Have you come across Lagrange polynomials? – Goran Malic Nov 10 at 9:29