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If $f(x),g(x)$ are real polynomials, then $f^2+g^2+1$ is a sum of two squares of polynomials. This easily follows from Fundamental Theorem of Algebra, but is there an argument avoiding it? What are fields for which such a statement holds?

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    $\begingroup$ $2x^2+1$ is clearly not the sum of two squares of $\mathbb{Q}$-coefficient polynomials (as $3$ is not the sum of two squares), so the statement does not hold over $\mathbb{Q}$. It does hold for real closed fields. $\endgroup$ – Vesselin Dimitrov Jun 18 '16 at 12:34
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    $\begingroup$ What is the easy proof with FTA? $\endgroup$ – Gerald Edgar Jun 18 '16 at 14:11
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    $\begingroup$ @GeraldEdgar Any non-negative polynomial $f(x)$ may have only real roots of even multiplicity and complex roots which are partitioned onto pairs of conjugates, thus $f(x)=C^2(x) (A(x)+iB(x))(A(x)-iB(x))=(CA)^2+(CB)^2$ for some real polynomials $C,A,B$. $\endgroup$ – Fedor Petrov Jun 18 '16 at 14:34
  • $\begingroup$ Yes for a field containing $\sqrt{-1}$. No for an ordered field containing a sum of 2 squares which isn't a square e.g. $\mathbb{R}(t): x^2+t^2+1$ is not a sum of 2 squares in $\mathbb{R}(t)[x]$. $\endgroup$ – David Lampert Jun 18 '16 at 16:07
  • $\begingroup$ @DavidLampert : $\sqrt{-1}\not\in\mathbb{R}$, yet the result holds... $\endgroup$ – Dima Pasechnik Jun 18 '16 at 18:43
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Theorem 4.2 from this paper says that in general you will need the coefficients of the two squares to be algebraic numbers of exponential, in $\deg (f^2+g^2+1)$, degrees. More precisely, if the Galois group of $f^2+g^2+1\in\mathbb{Q}[x]$ is the symmetric group $S_{d}$, then you will need algebraic numbers of degree exponential in $d$.

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