0021c: Article 13 (Pfister's 16-Square Identity)

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Pfister’s 16-Square Identity

By Tito Piezas III

Abstract: A non-bilinear identity allowed by Pfister’s theorem will be given for the form,

I. Introduction

II. Hurwitz’s Theorem

III. Pfister’s Theorem and the 16-Square Identity

IV. Some Properties

V. Acknowledgments

VI. LaTex Code

I. Introduction

Consider the Degen-Graves Eight-Square Identity,

If we choose to set all x_i and y_i equal to zero excepting i = {1, 2, 3, 4}, then what remains of the z_i are the variables in red and blue, and we get the Euler Four-Square Identity. If it is reduced even further to just i = {1, 2}, then what is left is the red square and this is the Brahmagupta-Fibonacci Two-Square Identity. One can then say that the Degen-Graves contains those two smaller ones.

If we go higher,

call this F₁₆, there can be a misconception that there is no sixteen-square identity. There is no bilinear identity for F₁₆, but there are non-bilinear ones which is allowed by Pfister’s theorem (1965). (Author’s note: The situation seems reminiscent of the unqualified statement that there is no formula for the general quintic. No formula just in the radicals, but if we expand our tool-set of functions, there is one using elliptic functions as shown by Hermite.)

There are various formulations but, in this author’s version, it will be given in semi-bilinear form: eight of the z_i are bilinear, while the other eight are non-bilinear. This 16-square identity then contains the Degen-Graves since setting the appropriate half of its {x_i, y_i} equal to zero reduces it to the latter. (By Shapiro’s theorem (1978), an identity for F₁₆ can have a maximum of nine bilinear z_i.)

II. Hurwitz’s Theorem

After Euler’s discovery of the four-square, and the eight-square by Degen, Graves, and Cayley, it was reasonable to search for a sixteen-square version. However, given an identity of form,

(x₁² + x₂² + x₃² + …+ x_m²) (y₁² + y₂² + y₃² + …+ y_m²) = z₁² + z₂² + z₃² + …+ z_n²

if m = n, and the z_i are to be bilinear in the x_i and y_i, then Hurwitz’s theorem (1898) states that the only possible identities of this sort are for n = {1, 2, 4, 8}, hence there are only four normed division algebras over the reals: the real numbers, complex numbers, quaternions, and octonions.

But there are ways we can evade Hurwitz’s theorem. First, we can drop the requirement that m = n. Given Lagrange’s polynomial identity,

this guarantees a bilinear identity for any m, at the cost that the number of terms on the RHS increase rapidly. For example, for m = 4, we already have seven terms on the RHS,

(x₁²+x₂²+x₃²+x₄²)(y₁²+y₂²+y₃²+y₄²) = z₁² + z₂² + z₃² + z₄² + z₅² + z₆² + z₇²

z₁ = x₁y₁+x₂y₂+x₃y₃+x₄y₄

z₂ = x₁y₂-x₂y₁; z₅ = x₂y₃-x₃y₂

z₃ = x₁y₃-x₃y₁; z₆ = x₂y₄-x₄y₂

z₄ = x₁y₄-x₄y₁; z₇ = x₃y₄-x₄y₃

and so on.

III. Pfister’s Theorem and the 16-Square Identity

The second way is to retain m = n, but to drop the bilinear requirement. An early non-bilinear 16-square identity was given in a paper (1966) by Zassenhaus and Eichhorn. Almost at the same time, Pfister (1965) established that, if the z_i are linear in only one variable and non-linear in the other, then there are m = n identities for ALL n = 2^k. Thus we have,

Pfister 4-Square

Note also the incidental fact that,

u₁²+u₂² = x₃²+x₄²

Pfister 8-Square:

Similarly to the previous,

u₁²+u₂²+u₃²+u₄² = x₅²+x₆²+x₇²+x₈²

Pfister 16-Square:

and,

Not surprisingly,

With all the variables z_i and u_i known, Mathematica can verify the 16-square identity in just 29 secs. One can recognize the three blue squares as the Brahmagupta-Fibonacci, Euler, and Degen-Graves, respectively. In fact, if the z_i array is to be divided into four “quadrants”, then each quadrant, save for a change of variables, has an identical template.

IV. Some Properties

The 16-square (and analogously for the two others) has several interesting properties:

First: If all x_i and y_i, excepting i = {1, 2, 3, … 8} are set equal to zero, then it reduces to the Degen-Graves.

Second: If the u_i are defined as u_i = x_i+8, then the identity remains true if,

(x₁x₉ + x₂x₁₀ + x₃x₁₁ +…+ x₈x₁₆) (y₁y₉ + y₂y₁₀ + y₃y₁₁ +…+ y₈y₁₆) = 0

which is one of the simplest possible conditions for a “quasi-bilinear” 16-square identity (at the cost of one variable being linearly dependent on the others).

Third: If any seven of {x₁, x₂, x₂, … x₈} are set equal to zero, then the denominator of the u_i vanishes, as well as the second powers in the numerators, and we have a fully bilinear form. If any seven of the y_i are also set to zero, then we get the nice form,

(p₁² + p₂² + p₃² + …+ p₉²) (q₁² + q₂² + q₃² + …+ q₉²) = z₁² + z₂² + z₃² + …+ z₁₆²

or a bilinear [9.9.16] which, for m = 9, is the minimum possible number of z_i . See Products of Sums of Squares by Shapiro for the limits for various m.

V. Acknowledgments

The author wishes to thank Keith Conrad for his paper, Pfister's Theorem on Sums of Squares which was the inspiration for this article. (I had long wanted to see how the 16-square identity looked like.)

VI. LaTex Code

For those who wish to verify the 16-square, the LaTex code for the z_i and u_i is given below to facilitate cut-and-paste onto software like Mathematica or Maple.

<math>\begin{align}

&^{z1 \,=\, x1 y1 - x2 y2 - x3 y3 - x4 y4 - x5 y5 - x6 y6 - x7 y7 - x8 y8 + u1 y9 - u2 y10 - u3 y11 - u4 y12 - u5 y13 - u6 y14 - u7 y15 - u8 y16}\\

&^{z2 \,=\, x2 y1 + x1 y2 + x4 y3 - x3 y4 + x6 y5 - x5 y6 - x8 y7 + x7 y8 + u2 y9 + u1 y10 + u4 y11 - u3 y12 + u6 y13 - u5 y14 - u8 y15 + u7 y16}\\

&^{z3 \,=\, x3 y1 - x4 y2 + x1 y3 + x2 y4 + x7 y5 + x8 y6 - x5 y7 - x6 y8 + u3 y9 - u4 y10 + u1 y11 + u2 y12 + u7 y13 + u8 y14 - u5 y15 - u6 y16}\\

&^{z4 \,=\, x4 y1 + x3 y2 - x2 y3 + x1 y4 + x8 y5 - x7 y6 + x6 y7 - x5 y8 + u4 y9 + u3 y10 - u2 y11 + u1 y12 + u8 y13 - u7 y14 + u6 y15 - u5 y16}\\

&^{z5 \,=\, x5 y1 - x6 y2 - x7 y3 - x8 y4 + x1 y5 + x2 y6 + x3 y7 + x4 y8 + u5 y9 - u6 y10 - u7 y11 - u8 y12 + u1 y13 + u2 y14 + u3 y15 + u4 y16}\\

&^{z6 \,=\, x6 y1 + x5 y2 - x8 y3 + x7 y4 - x2 y5 + x1 y6 - x4 y7 + x3 y8 + u6 y9 + u5 y10 - u8 y11 + u7 y12 - u2 y13 + u1 y14 - u4 y15 + u3 y16}\\

&^{z7 \,=\, x7 y1 + x8 y2 + x5 y3 - x6 y4 - x3 y5 + x4 y6 + x1 y7 - x2 y8 + u7 y9 + u8 y10 + u5 y11 - u6 y12 - u3 y13 + u4 y14 + u1 y15 - u2 y16}\\

&^{z8 \,=\, x8 y1 - x7 y2 + x6 y3 + x5 y4 - x4 y5 - x3 y6 + x2 y7 + x1 y8 + u8 y9 - u7 y10 + u6 y11 + u5 y12 - u4 y13 - u3 y14 + u2 y15 + u1 y16}\\

&^{z9 \,=\, x9 y1 - x10 y2 - x11 y3 - x12 y4 - x13 y5 - x14 y6 - x15 y7 - x16 y8 + x1 y9 - x2 y10 - x3 y11 - x4 y12 - x5 y13 - x6 y14 - x7 y15 - x8 y16}\\

&^{z10 \,=\, x10 y1 + x9 y2 + x12 y3 - x11 y4 + x14 y5 - x13 y6 - x16 y7 + x15 y8 + x2 y9 + x1 y10 + x4 y11 - x3 y12 + x6 y13 - x5 y14 - x8 y15 + x7 y16}\\

&^{z11 \,=\, x11 y1 - x12 y2 + x9 y3 + x10 y4 + x15 y5 + x16 y6 - x13 y7 - x14 y8 + x3 y9 - x4 y10 + x1 y11 + x2 y12 + x7 y13 + x8 y14 - x5 y15 - x6 y16}\\

&^{z12 \,=\, x12 y1 + x11 y2 - x10 y3 + x9 y4 + x16 y5 - x15 y6 + x14 y7 - x13 y8 + x4 y9 + x3 y10 - x2 y11 + x1 y12 + x8 y13 - x7 y14 + x6 y15 - x5 y16}\\

&^{z13 \,=\, x13 y1 - x14 y2 - x15 y3 - x16 y4 + x9 y5 + x10 y6 + x11 y7 + x12 y8 + x5 y9 - x6 y10 - x7 y11 - x8 y12 + x1 y13 + x2 y14 + x3 y15 + x4 y16}\\

&^{z14 \,=\, x14 y1 + x13 y2 - x16 y3 + x15 y4 - x10 y5 + x9 y6 - x12 y7 + x11 y8 + x6 y9 + x5 y10 - x8 y11 + x7 y12 - x2 y13 + x1 y14 - x4 y15 + x3 y16}\\

&^{z15 \,=\, x15 y1 + x16 y2 + x13 y3 - x14 y4 - x11 y5 + x12 y6 + x9 y7 - x10 y8 + x7 y9 + x8 y10 + x5 y11 - x6 y12 - x3 y13 + x4 y14 + x1 y15 - x2 y16}\\

&^{z16 \,=\, x16 y1 - x15 y2 + x14 y3 + x13 y4 - x12 y5 - x11 y6 + x10 y7 + x9 y8 + x8 y9 - x7 y10 + x6 y11 + x5 y12 - x4 y13 - x3 y14 + x2 y15 + x1 y16}\\

\end{align}</math>

<math>\begin{align}

&^{u1 \,=\,} \tfrac{(-x1^2+x2^2+x3^2+x4^2+x5^2+x6^2+x7^2+x8^2)x9 - 2x1(0 x1 x9 +x2 x10 +x3 x11 +x4 x12 +x5 x13 +x6 x14 +x7 x15 +x8 x16)}{d}\\

&^{u2 \,=\,} \tfrac{(x1^2-x2^2+x3^2+x4^2+x5^2+x6^2+x7^2+x8^2)x10 - 2x2(x1 x9 + 0 x2 x10 +x3 x11 +x4 x12 +x5 x13 +x6 x14 +x7 x15 +x8 x16)}{d}\\

&^{u3 \,=\,} \tfrac{(x1^2+x2^2-x3^2+x4^2+x5^2+x6^2+x7^2+x8^2)x11 - 2x3(x1 x9 +x2 x10 + 0 x3 x11 +x4 x12 +x5 x13 +x6 x14 +x7 x15 +x8 x16)}{d}\\

\vdots\\

&^{u8 \,=\,} \tfrac{(x1^2+x2^2+x3^2+x4^2+x5^2+x6^2+x7^2-x8^2)x16 - 2x8(x1 x9 +x2 x10 +x3 x11 +x4 x12 +x5 x13 +x6 x14 +x7 x15 + 0 x8 x16)}{d}\\

\,&^{d \,=\, x1^2+x2^2+x3^2+x4^2+x5^2+x6^2+x7^2+x8^2}

\end{align}</math>

-- End --

© Tito Piezas III, Jan 2012 (Modified Feb 2012)

You can email author at tpiezas@gmail.com.

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