9
$\begingroup$

Consider this 5-Square Identity,

$(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)^2 (y_1^2+y_2^2+y_3^2+y_4^2+y_5^2) = z_1^2+z_2^2+z_3^2+z_4^2+z_5^2$

where,

$\begin{align} z_1 &= (-x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)y_1 - 2x_1(0x_1 y_1+x_2 y_2+x_3 y_3+x_4 y_4 + x_5 y_5)\\ z_2 &= (x_1^2-x_2^2+x_3^2+x_4^2+x_5^2)y_2 - 2x_2(x_1 y_1+0x_2 y_2+x_3 y_3+x_4 y_4 + x_5 y_5)\\ z_3 &= (x_1^2+x_2^2-x_3^2+x_4^2+x_5^2)y_3 - 2x_3(x_1 y_1+x_2 y_2+0x_3 y_3+x_4 y_4 + x_5 y_5)\\ z_4 &= (x_1^2+x_2^2+x_3^2-x_4^2+x_5^2)y_4 - 2x_4(x_1 y_1+x_2 y_2+x_3 y_3+0x_4 y_4 + x_5 y_5)\\ z_5 &= (x_1^2+x_2^2+x_3^2+x_4^2-x_5^2)y_5 - 2x_5(x_1 y_1+x_2 y_2+x_3 y_3+x_4 y_4 + 0x_5 y_5) \end{align}$

The pattern is easily seen for,

$(x_1^2+x_2^2 + \dots + x_n^2)^2 (y_1^2+y_2^2 + \dots + y_n^2) = z_1^2+z_2^2 + \dots + z_n^2$

The case n = 4 is used in Pfister’s 8-square Identity. How to prove the pattern indeed holds true for ALL positive integer n?

$\endgroup$
3
  • $\begingroup$ Is there a square wrong on $x_1^2+\dots+x_5^2$ in the title and the first equation? $\endgroup$
    – emiliocba
    Jan 30, 2012 at 19:44
  • $\begingroup$ Look at first term. Get $(\sum x_i^2)-2x_1^2$. Also, have in second summand of first term, $-2x_1((\sum x_iy_i)-x_1y_1)$. But still unpleasant! $\endgroup$
    – André Nicolas
    Jan 30, 2012 at 19:49
  • $\begingroup$ @emiliocba: No, I've checked it with Mathematica and the 5-Square identity holds true. $\endgroup$
    – Tito Piezas III
    Jan 30, 2012 at 20:02

1 Answer 1

Reset to default
6
$\begingroup$

We can write \begin{align*} z_k&=y_k\left(\sum_ix_i^2-2x_k^2\right)-2x_k\sum_{i\neq k}x_iy_i\\ &=y_k\sum_ix_i^2-2x_k\left(\sum_{i\neq k}x_iy_i+x_ky_k\right)\\ &=y_k\sum_ix_i^2-2x_k\sum_ix_iy_i \end{align*} so hopping we are working in a commutative ring $$z_k^2=y_k^2\left(\sum_ix_i^2\right)^2-4x_ky_k\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)+4x_k^2\left(\sum_ix_iy_i\right)^2$$ and finally \begin{align*} \sum_{k=1}^nz_k^2&=\sum_{k=1}^ny_k^2\left(\sum_ix_i^2\right)^2-4x_ky_k\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)+4x_k^2\left(\sum_ix_iy_i\right)^2\\ &=\left(\sum_ix_i^2\right)^2\sum_{k=1}^ny_k^2-4\left(\sum_kx_ky_k\right)\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)\\ &+4\left(\sum_kx_k^2\right)\left(\sum_ix_iy_i\right)^2\\ &=\left(\sum_ix_i^2\right)^2\sum_{k=1}^ny_k^2. \end{align*}

$\endgroup$
5
  • $\begingroup$ strictly speaking, is it true that: $$\sum_{k=1}^{n}-4\left(\sum_{k} x_{k}y_{k}\right)\left(\sum_{i} x_{i}^2\right) \left(\sum_{i} x_iy_i\right)+4\left(\sum_{k}x_{k}^2\right)\left(\sum_{i} x_{i}y_{i}\right)^2=0$$??????? I am not saying you are incorrect, I am simply saying that it looks bizarre to me. Do you have any way of clarifying this? My issue lies in the fact that you are cancelling summations which have similar terms but different indexes. $\endgroup$
    – 000
    Jan 30, 2012 at 20:55
  • $\begingroup$ @user22144: in the equality you write, there should not be $\sum_{k=1}^n$. The last step of my computation is in fact a consequence of the fact that $\sum_{i=1}^nc_i=\sum_{j=1}^nc_j$. $\endgroup$
    – Davide Giraudo
    Jan 30, 2012 at 21:02
  • $\begingroup$ Dear @Davide: Thanks so much! When I first stumbled upon the general form in the context of "Pfister's 8-Square Identity", I assumed it was only for n = 2^m. But a little experimentation with Mathematica showed it was was more than that. It is good to know it is in fact valid for all n. Thanks again. $\endgroup$
    – Tito Piezas III
    Jan 31, 2012 at 5:42
  • $\begingroup$ @DavideGiraudo, my mistake! I had misinterpreted the sum. No wonder it looked so bizarre. $\endgroup$
    – 000
    Jan 31, 2012 at 20:34
  • $\begingroup$ Davide, you may have interests in this: math.stackexchange.com/questions/751167/… :0) $\endgroup$
    – annie marie cœur
    Apr 12, 2014 at 22:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .