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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?

  • 0
    @emiliocba: No, I've checked it with Mathematica and the 5-Square identity holds true.2012-01-30

1 Answers 1

5

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*}

  • 0
    Davide, you may have interests in this: http://math.stackexchange.com/questions/751167/counting-problem-very-interesting-modular-n-algebraic-eqs-for-combinatorics?lq=1 :0)2014-04-12