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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.

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    $\begingroup$ what did you try? for e.g. what happens to the leading coefficients of $p,q$ when squared and added? $\endgroup$
    – Macavity
    Dec 8, 2020 at 20:11
  • $\begingroup$ If degree of $p$ is at least degree of $q$, and that degree is $k$ look at coefficients of $t^{2k}$. $\endgroup$ Dec 8, 2020 at 20:12
  • $\begingroup$ @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? $\endgroup$
    – Figurinha
    Dec 8, 2020 at 20:55

2 Answers 2

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

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    $\begingroup$ $p+iq$ constant does not imply $p$ and $q$ are constant, but $p+iq$ and $p-iq$ both constant does. $\endgroup$ Dec 8, 2020 at 20:45
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    $\begingroup$ @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 $\mathbb R$. $\endgroup$ Dec 8, 2020 at 20:54
  • $\begingroup$ 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. $\endgroup$ Dec 8, 2020 at 21:07
  • $\begingroup$ I added in the question that the problem is about real polynomials, but the extra cases are also interesting! $\endgroup$
    – Figurinha
    Dec 8, 2020 at 21:09
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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.

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

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