Let's show that the Motzkin polynomial cannot be expressed as the sum of squares. Assume for the sake of contradiction, that
where are all real constants. By comparing the above expression to Motzkin's polynomial, we see that some of the coefficients in equal zero, because Motzkin's polynomial does not contain some terms in
Firstly, Motzkin's polynomial does not have , so we can eliminate the constants and , that is, .
So now
Additionally, Motzkin's polynomial does not have the terms , so we know that .
So now
Likewise, there is no term in the Motzkin polynomial, which means that , hence each . Finally,there is no term, so the coefficient of of is zero, so and each
Therefore, we are left with where are constants.
Following this procedure, we see that the coefficient of the term in the Motzkin polynomial is -3. This means that
But if we take a look at , we see that there is no term in the Motzkin polynomial, so . But we found above that , so , hence each . Therefore,
Since the sum of all the terms cannot be negative, this is a contradiction.
Hence we conclude that the Motzkin polynomial, , is an example of a non-negative polynomial of two variables that cannot be written as the sum of squares.
Previously, we showed that the Motzkin polynomial is non-negative using the AM-GM inequality. This polynomial caught our interest, because it is non-negative in , but cannot be expressed as the sum of two squares of polynomials. The Motzkin polynomial illustrates that although a non-negative polynomial in one variable can always be expressed as the sum of squares, this is not necessarily the case for a polynomial in two variables.
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