No polynomials of degree > 0 are such that $p(t)^2 + q(t)^2 =1$

Question

I'm trying to prove here that ellipses can't be represented by a parametric polynomial and got stuck on this part of the problem:

No real polynomials of degree $> 0$ are such that $$p(t)^2 + q(t)^2 =1$$

I want to prove the above statement.

Answer 1

Over $\mathbb C$ you would have $(p+iq)(p-iq)=1$, which is a factorisation of a constant polynomial $1$ in $\mathbb C[x]$. This implies that $p+iq$ and $p-iq$ are constant polynomial themselves, which implies that $p$ and $q$ are both constant.

(The last conclusion follows directly if $p$ and $q$ are real polynomials, however we can see it is valid even if $p$ and $q$ were complex polynomials to start with, using: $p=\frac{1}{2}((p+iq)+(p-iq))$ and $q=\frac{1}{2i}((p+iq)-(p-iq))$)

Answer 2

Real polynomials that are not constants go to infinity at infinity (just factor out the term of higher degree, other terms are negligible).

Squaring them make them both positive, i.e. $p(t)^2\to+\infty$ and $q(t)^2\to+\infty$ so their sum cannot be bounded.