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

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.

• what did you try? for e.g. what happens to the leading coefficients of $p,q$ when squared and added? Dec 8, 2020 at 20:11
• If degree of $p$ is at least degree of $q$, and that degree is $k$ look at coefficients of $t^{2k}$. Dec 8, 2020 at 20:12
• @Macavity if the coefficients are different from zero they will always be positive! wich contradicts the fact that the sum of both is constant... Am I right? Dec 8, 2020 at 20:55

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 $$q=\frac{1}{2i}((p+iq)-(p-iq))$$)

• $p+iq$ constant does not imply $p$ and $q$ are constant, but $p+iq$ and $p-iq$ both constant does. Dec 8, 2020 at 20:45
• @RobertIsrael I will fix it. This is necessary in $\mathbb C$, though I assumed (based on the linked post) that the OP was working in $\mathbb R$. Dec 8, 2020 at 20:54
• Come to think about it, I think this is also valid in any field of characteristic $\ne 2$, because either there is already an "$i$" in the field (i.e. a solution of $x^2+1=0$), or the field can be extended with a solution of that equation just as we did for $\mathbb R$ - the only problem is the division by $2$ in the last two formulas. If the field is of characteristic $2$, then $(x+1)^2+x^2=1$ so the statement is false. Dec 8, 2020 at 21:07
• I added in the question that the problem is about real polynomials, but the extra cases are also interesting! Dec 8, 2020 at 21:09

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.

• Edited to remove the coercive word, it was misleading, sign is important here.
– zwim
Dec 8, 2020 at 21:42