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Given a polynomial of degree $2n$ over $\mathbb{Q}$, how to represent it as a linear combination (with rational coefficients) of squares of polynomials of degree at most $n$ over $\mathbb{Q}$ such that the number of polynomials is minimal?

In particular, when it is possible to represent a given polynomial of degree $2n$ as a linear combination of two squares of polynomials of degree at most $n$?

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    $\begingroup$ When it is reducible with $f=uv$, and degree of $u$, $v$ are $n$, then $p=((u+v)/2)^2-((u-v)/2)^2$. But, not sure how to go for other cases. $\endgroup$ Jun 2, 2013 at 23:44

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Ok, here's how to write a polynomial $f$ of degree $2n$ as a linear combination of three squares. We can assume $f$ is monic, since everything can be scaled at the end. Now, complete squares, starting from the top degree term. So we can write $f = g^2 + h$, where $g$ is monic of degree $n$, and $h$ has degree at most $n-1$. Then write $h = \frac{1}{4}(h+1)^2 - \frac{1}{4}(h-1)^2$ to finish it off. Three squares is optimal: a little bit of work shows that $x^4 + x + 1$ cannot be written as a linear combination of two squares (the same holds for a generic quartic polynomial). I guess this doesn't fully answer your second question of how to characterize linear combinations of two squares of degree $n$.

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  • $\begingroup$ Why does $h$ have degree at most $n-1$? I know that it should be $\leq 2n-1$, but how do you get $n-1$? $\endgroup$ Jun 5, 2013 at 15:14
  • $\begingroup$ As an example, $x^4 + 2x^3 + 3x^2 + 5x + 7 = (x^2 + x)^2 + 2x^2 + 5x + 7 = (x^2 + x + 1)^2 + 3x + 6$. Think of this as taking square roots near the place $x = \infty$ on the projective line. $\endgroup$ Jun 5, 2013 at 16:33
  • $\begingroup$ Thanks, I got it now. It can also be done by induction on degree. $\endgroup$ Jun 5, 2013 at 18:40
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Here is a comment relating the question to representations of sums of five squares of rational polynomials (sorry, too long for the comment field):

Suppose that $f$ is a linear combination of two squares of polynomials, i.e., $f=u^2-rv^2$ for some $r\in \mathbb{Q}$, and $u,v\in \mathbb{Q}[x]$. If $r\le 0$, then $f$ need to be nonnegative. As $-r$ is the sum of four rational squares, we can write $f$ as the sum of five squares of rational polynomials. Conversely, if $f$ is nonnegative, then it is known that $f$ can be written as the sum of five squares of rational polynomials, $f=u_1^2+\cdots +u_5^2$. This is due to Pourchet, 1971, "Sur la représentation en somme de carrés des polynômes à une indéterminée sur un corps de nombres alge ́briques". This result is best possible as far as the number of squares needed for such representations is concerned.

Edit: The result of Pourchet does not imply that $f$ may not be a linear combination of two squares of rational polynomials. Consider the example $f(x)=x^2+x+4$. Since $4ac-b^2=15$ cannot be represented as a sum of three rational squares, the polynomial $f$ cannot be represented as the sum of squares of $4$ rational polynomials. However, $$ f(x)=x^2+x+4=\left( \\frac{x-7}{4}\right)^2+\frac{15}{16}(x+1)^2. $$

If $r>0$, then $f$ must be reducible over $\mathbb{R}$, as $f=(u-\sqrt{r}v)(u+\sqrt{r}v)$ (see the comment of i707107).

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    $\begingroup$ If I'm understanding Pourchet's result correctly, it seems to me that one should be able to represent any polynomial $f$ of even degree as a linear combination of six squares. We can assume $f$ is monic, then if you add a sufficiently large constant $c^2$ (with $c$ rational) to $f$, it will become non-negative everywhere: just take $c^2$ larger than the minimum value of $f$, noting $f$ goes to $\infty$ as the argument goes to $\pm \infty$. So by Pourchet, $f + c^2$ is a sum of at most five squares, and $f$ is a linear combination (with $\pm 1$ coeffs) of six squares. I doubt this is optimal! $\endgroup$ Jun 4, 2013 at 3:24
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If $f = u^2- r v^2$, then $f=(u-\sqrt{r}v)(u+\sqrt{r}v)$, so $f$ is reducible over $\mathbb Q(\sqrt{r})$. This provides a strong condition on the Galois group of $f$ - an index $2$ subgroup must act nontransitively on the roots. In other words, the Galois group must be a subgroup of the wreath product of $\mathbb Z/2$ and $S_n$.

Any polynomial with Galois group full$S_{2n}$, $n>1$, violates this condition.

On the other hand, a polynomial that does satisfy this Galois group condition can be written in this way. Over $\mathbb Q(\sqrt{r})$, it splits into two factors, say $a + b\sqrt{r}$ and $c+d\sqrt{r}$. But these two factors must be Galois conjugates, so $c=a $ and $d=-b$. So $f=a^2-r b^2$. (This is all up to a constant factor)

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Somewhat related topic is the Waring problem for polynomials, namely to estimate the number $k(n)$ such that every polynomial $P(x)$ over $\mathbb{C}[x]$ can be represented as a sum $P(x)=\sum_{k=1}^{k(n)}Q_k(x)^{n}.$ Clearly, $k(2)=2$ (just take $P(x)=1/4(P(x)+1)^2-1/4(P(x)-1)^2.$ This particular example shows that the number of polynomials needed to represent over $Q[x]$ can be larger than $k(n).$ In general, for $n\ge 3$ Newman and Slater ($1979$) proved the bound $k(n)\le n< k^2(n)-k(n).$

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