## A Collection of Algebraic Identities

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

 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 xi and yi equal to zero excepting i = {1, 2, 3, 4}, then what remains of the zi 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 zi 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 {xi, yi} equal to zero reduces it to the latter.  (By Shapiro’s theorem (1978), an identity for F16 can have a maximum of nine bilinear zi.)        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 + …+ xm2) (y12 + y22 + y32 + …+ ym2) = z12 + z22 + z32 + …+ zn2   if m = n, and the zi are to be bilinear in the xi and yi, 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 zi are linear in only one variable and non-linear in the other, then there are m = n identities for ALL n = 2k. 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 xi and yi, excepting i = {1, 2, 3, … 8} are set equal to zero, then it reduces to the Degen-Graves.   Second:  If the ui are defined as ui = xi+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 ui vanishes, as well as the second powers in the numerators, and we have a fully bilinear form.  If any seven of the yi 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 zi . 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 zi and ui is given below to facilitate cut-and-paste onto software like Mathematica or Maple.   \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}   \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}      -- End --       © Tito Piezas III, Jan 2012 (Modified Feb 2012) You can email author at tpiezas@gmail.com.       ◄