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.

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