# [Number Theory] Using Brahmagupta Identity and FTA to show Non-Negative Univariate Polynomial is SOS

Let's use the Fundamental Theorem Algebra (FTA) and the Brahmagupta Fibonacci identity to prove that every non-negative polynnomial $$f(x) \in \mathbb{R}\left[ x \right]$$ may be written as a sum of squares of polynomials.

Let $$p \in \mathbb{R}[x] ~ \forall x \in \mathbb{R}$$ and $$p(x) \geq 0$$. By the Fundamental Theorem of Algebra, a degree $$n$$ polynomial $$p$$ has exactly $$n$$ complex roots, and our polynomial $$p$$ is of the form constant times a product of positive quadratics. The graph of the polynomial lies entirely above the x-axis, or touches it. So in the first case, the roots of p are all complex, whereas in the latter, $$p$$ has real roots with even multiplicity. Hence our polynomial is of the form

$p = a~f_1^2f_2^2 \cdots f_n^2g_1 \cdots g_m,$

where $$a$$ is a real coefficient, the $$f_i$$'s have degree 1, and $$g_j$$'s have degree 2 and no real roots. We also note that by the complex conjugate root theorem, the roots of $$p$$ come in complex conjugate pairs. Also, $$p(x)$$ does not intersect the x-axis, so the degree of $$p$$ must be even.

A monic positive quadratic $$g_j$$, with no real roots, can be written as $$(x+a)^2+b^2$$ by completing the square. From here, an expression of $$p$$ as a sum of squares follows from the Brahmagupta-Fibonacci identity; the product of $$g_j$$'s, hence a product of the sums of squares, can be written as the sum of squares. For example, the product of two quadratics

$((x+a)^2+b^2)((x+c)^2+d^2)$

$= ((x+a)(x+c) + b^2d^2) + ((x+a)^2d^2-b^2(x+c)).$

If the degree of $$p$$ is 2, then

$p = (x-(a+ib))(x-(a-ib))$

$= x^2-2ax+a^2+b^2 = (x-a)^2+b^2$

where $$a,b \in R[x],$$ hence the sum of two squares.

If degree of $$p$$ is greater than or equal to 4, then we can write $$p$$ as the product of quadratics. By applying the Brahmagupta-Fibonacci identity as needed, the product of many sums of squares can be written as a sum of many squares.

Since the degree of $$p$$ is not odd, we have shown that a non-negative univariate polynomial is a sum of squares of polynomials. $$\blacksquare$$

Note that in the case that $$p=a$$, where $$a\in \mathbb{R}$$, $$p$$ is itself a square.

Note by Tasha Kim
1 month ago

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