We are familiar with Hurwitz’s theorem which implies there is only the Fibonacci 2-Square, Euler 4-Square, Degen 8-Square, and no more. However, if we relax conditions and allow for rational expressions, then Pfister's theorem states that similar identities are possible for ALL $2^n$ squares. His 4-square version is:

$\begin{align}&(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)=\\ &(a_1 b_4 + a_2 b_3 + a_3 b_2 + a_4 b_1)^2 +\\ &(a_1 b_3 - a_2 b_4 + a_3 b_1 - a_4 b_2)^2 +\\ &\left(a_1 b_2 + a_2 b_1 + \frac{a_3 (b_1^2b_4-2b_1b_2b_3-b_2^2b_4)}{b_1^2+b_2^2} - \frac{a_4 (b_1^2b_3+2b_1b_2b_4-b_2^2b_3)}{b_1^2+b_2^2}\right)^2+\\ &\left(a_1 b_1 - a_2 b_2 - \frac{a_4 (b_1^2b_4-2b_1b_2b_3-b_2^2b_4)}{b_1^2+b_2^2} - \frac{a_3 (b_1^2b_3+2b_1b_2b_4-b_2^2b_3)}{b_1^2+b_2^2}\right)^2 \end{align}$

Question: What does the Pfister 8-square version look like? (And, if you have the strength, can you also give the 16-square version?) Here is K. Conrad's pdf file which describes the general method, but I can’t make heads or tails out of it.

$\color{red}{\text{Attention}}$ (Feb. 16): Someone is trying to delete Wikipedia's article on Degen's Eight-Square Identity simply because he finds it uninteresting. Please vote to keep.


2 Answers 2


Persistence pays off. I really wanted to see Pfister’s 8-Square Identity (distinct from Degen's version) and, since no one was answering my question, I took another look at K. Conrad’s paper and managed, with some heuristics, to find the identity myself. Without further ado,

$\begin{align} &(x_1^2\,+\,x_2^2\,+\,x_3^2\,+\,x_4^2\,+\,x_5^2\,+\,x_6^2\,+\,x_7^2\,+\,x_8^2)(y_1^2\,+\,y_2^2\,+\,y_3^2\,+\,y_4^2\,+\,y_5^2\,+\,y_6^2\,+\,y_7^2\,+\,y_8^2)\\ &=z_1^2\,+\,z_2^2\,+\,z_3^2\,+\,z_4^2\,+\,z_5^2\,+\,z_6^2\,+\,z_7^2\,+\,z_8^2 \end{align}$


$\begin{align} z_1 &= x_1 y_1 - x_2 y_2 - x_3 y_3 - x_4 y_4 + u_1 y_5 - u_2 y_6 - u_3 y_7 - u_4 y_8\\ z_2 &= x_2 y_1 + x_1 y_2 + x_4 y_3 - x_3 y_4 + u_2 y_5 + u_1 y_6 + u_4 y_7 - u_3 y_8\\ z_3 &= x_3 y_1 - x_4 y_2 + x_1 y_3 + x_2 y_4 + u_3 y_5 - u_4 y_6 + u_1 y_7 + u_2 y_8\\ z_4 &= x_4 y_1 + x_3 y_2 - x_2 y_3 + x_1 y_4 + u_4 y_5 + u_3 y_6 - u_2 y_7 + u_1 y_8\\ z_5 &= x_5 y_1 - x_6 y_2 - x_7 y_3 - x_8 y_4 + x_1 y_5 - x_2 y_6 - x_3 y_7 - x_4 y_8\\ z_6 &= x_6 y_1 + x_5 y_2 + x_8 y_3 - x_7 y_4 + x_2 y_5 + x_1 y_6 + x_4 y_7 - x_3 y_8\\ z_7 &= x_7 y_1 - x_8 y_2 + x_5 y_3 + x_6 y_4 + x_3 y_5 - x_4 y_6 + x_1 y_7 + x_2 y_8\\ z_8 &= x_8 y_1 + x_7 y_2 - x_6 y_3 + x_5 y_4 + x_4 y_5 + x_3 y_6 - x_2 y_7 + x_1 y_8 \end{align}$


$\begin{align} u_1 &= \frac{(-x_1^2+x_2^2+x_3^2+x_4^2)x_5 - 2x_1(0x_1 x_5+x_2 x_6+x_3 x_7+x_4 x_8)}{d}\\ u_2 &= \frac{(x_1^2-x_2^2+x_3^2+x_4^2)x_6 - 2x_2(x_1 x_5+0x_2 x_6+x_3 x_7+x_4 x_8)}{d}\\ u_3 &= \frac{(x_1^2+x_2^2-x_3^2+x_4^2)x_7 - 2x_3(x_1 x_5+x_2 x_6+0x_3 x_7+x_4 x_8)}{d}\\ u_4 &= \frac{(x_1^2+x_2^2+x_3^2-x_4^2)x_8 - 2x_4(x_1 x_5+x_2 x_6+x_3 x_7+0x_4 x_8)}{d}\\ d &=x_1^2+x_2^2+x_3^2+x_4^2 \end{align}$

I’ve verified it with Mathematica and it holds true. Note also the nice sub-identity,

$u_1^2+u_2^2+ u_3^2+ u_4^2 = x_5^2+x_6^2+ x_7^2+x_8^2$

P.S. By Pfister’s Theorem, the 16-square version is analogous.


Here's Maple code for getting a 16-square version using the method Conrad outlines. The result is too big to show here (the equation eq has length 4808351).

C2:= < < c[1] | c[2] >,<-c[2] | c[1]> >:

c2:= c[1]^2 + c[2]^2:

C2b:= subs(seq(c[i]=c[i+2],i=1..2),C2):

c4:= add(c[i]^2,i=1..4):

C4:= < < C2|C2b >,< -C2 . C2b^%T . C2/c2 | C2 > >:

C4b:= subs(seq(c[i]=c[i+4],i=1..4),C4):

c8:= add(c[i]^2,i=1..8):

C8:= < < C4|C4b >,< -C4 . C4b^%T . C4/c4 | C4 > >:

C8b:= subs(seq(c[i]=c[i+8],i=1..8),C8):

C16:= < < C8|C8b >,< -C8 . C8b^%T . C8/c8 | C8 > >:

XY:= subs(c=x, C16) . subs(c=y, C16):

for i to 16 do z[i] := XY[1, i] end do:

To verify:

eq:= add(x[i]^2, i=1..16) * add(y[i]^2, i=1..16) = add(z[i]^2, i=1..16):



  • $\begingroup$ Dear @Robert: Thank you so much! Are the individual z_i's manageable in size? I assume z_1 is of the form $z_1 = x_1 y_1 - x_2 y_2 - \dots - x_8 y_8 + u_1 y_9 + u_2 y_{10} + \dots + u_8 y_{16}$. Does the $u_1^2 + u_2^2 + \dots + u_8^2$ add up to anything special like in the 8-square version? $\endgroup$ Jan 31, 2012 at 5:29
  • $\begingroup$ Dear @Robert: Inspired by your result that it was doable, I re-visited my own methods and also found the 16-square identity. Yes, the individual z_i's are manageable in size and I'll put them in my website soon. And the $u_1^2+u_2^2 + \dots + u_8^2$ add up to $x_9^2+x_{10}^2 + \dots + x_{16}^2$ analogous to the one for 8-squares. $\endgroup$ Jan 31, 2012 at 9:18
  • $\begingroup$ Actually each $z_i$ is of the form $\sum_{i=1}^{16} u_{ij} x_i$ where the $u_{ij}$ are functions of the $y$'s. $z_{15} = x_{{1}}y_{{15}}-x_{{2}}y_{{16}}+x_{{3}}y_{{13}}-x_{{4}}y_{{14}}+x_{{5} }y_{{11}}-x_{{6}}y_{{12}}+x_{{7}}y_{{9}}-x_{{8}}y_{{10}}+x_{{9}}y_{{7} }-x_{{10}}y_{{8}}+x_{{11}}y_{{5}}-x_{{12}}y_{{6}}+x_{{13}}y_{{3}}-x_{{ 14}}y_{{4}}+x_{{15}}y_{{1}}-x_{{16}}y_{{2}}$ and $z_{16} = \sum_{i=1}^{15} x_i y_{17-i}$. $z_{13}$ and $z_{14}$ are not too bad either, but rather big to show in a comment. $\endgroup$ Jan 31, 2012 at 9:25
  • $\begingroup$ Sorry, that should have been $z_i = \sum_{j=1}^{16} u_{ij} x_j$. And it seems that in each case, $\sum_{j=1}^{16} u_{ij}^2 = \sum_{j=1}^{16} y_j^2$ $\endgroup$ Jan 31, 2012 at 9:37
  • $\begingroup$ Yes, eight of the $z_i$ have no denominators and are simple in form. The other eight, however, have denominators but can nonetheless be simplified. For those who wish to see a subjectively cute version of Pfister's 16-square identity given explicitly, see sites.google.com/site/tpiezas/0021c $\endgroup$ Jan 31, 2012 at 14:02

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