Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

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):

testeq(eq);

true

share|improve this answer
    
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? –  Tito Piezas III Jan 31 '12 at 5:29
    
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. –  Tito Piezas III Jan 31 '12 at 9:18
    
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. –  Robert Israel Jan 31 '12 at 9:25
    
Sorry, that should have been $z_i = \sum_{j=1}^{16} u_{ij} x_j$. And it seems that in each case, $\sum_{j=1}^{16} u_{ij}^2 = \sum_{j=1}^{16} y_j^2$ –  Robert Israel Jan 31 '12 at 9:37
    
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 –  Tito Piezas III Jan 31 '12 at 14:02
show 3 more comments

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,

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

where,

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

and,

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

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.