# Finding unknowns in terms of polynomials

Consider the following equation $$(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2})(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}+y_{5}^{2}) = z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}+z_{5}^{2}$$

How would we find the $z_i$ in terms of polynomials of $x_i,y_i$ (integer coefficients) such that the above holds? Just simply expand everything out and solve for $z_i$?

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I think you wanted to mean this. – Abhra Abir Kundu Jun 11 '13 at 18:43
Why the downvotes? This isn't the best-expressed question, but it's no worse than many on the site, and it asks an interesting question. – Steven Stadnicki Jun 11 '13 at 18:48
@AbhraAbirKundu: I meant what I said. Edit it back – grayguy Jun 11 '13 at 18:58
There is a formula giving the product of two sums of 5 squares as a sum of 11 squares, with $z_1=x_1y_1+...+x_5y_5$ and the other ten $z$ as the values of $x_iy_j-x_jy_i$ where $1 \le i < j \le 5$. – coffeemath Jun 12 '13 at 13:23

I am pretty confident that there is no such formula, by the same sort of ideas that show there are division algebras only in dimensions 1,2,4,8. This problem is quite easy in dimension 4, just use quaternion multiplication.

Well, it may take me a while to find a really specific reference on nonexistence. For the moment, pages 127-131 in Introduction to Quadratic Forms over Fields by T. Y. Lam.

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The left side is just a constant-no reason to have all those terms. You have five unknowns and one equation. As long as the left side is greater than $0$ you can choose $z_1,z_2,z_3,z_4$ arbitrarily as long as the sum of their squares is not too large, then find two values for $z_5$.

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OP wants integers, but in the same spirit as your answer, if $n$ is the left side, one can use LaGrange's four-square theorem and one var $0$ on the right. – coffeemath Jun 11 '13 at 19:03
@coffeemath: Abhra Abir Kundu thought OP wanted integers, but it seems not. But this is a good approach in that case. – Ross Millikan Jun 11 '13 at 19:11
@RossMillikan I think he wanted something like that so I decided to edit the question. But I think I might have been mistaken.Now I have asked him to reply if my answer is not what he wanted so that I can delete it. – Abhra Abir Kundu Jun 11 '13 at 19:25
@coffeemath I have edited it back. – Abhra Abir Kundu Jun 11 '13 at 19:46
@AbhraAbirKundu: Thanks -- now it is clearer what the OP was going for, something like complex multiplication but in the five real variable case... – coffeemath Jun 12 '13 at 0:53

This might help $(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=(ac-bd)^2+(ad+bc)^2$

$(x^2+y^2+z^2+w^2)(a^2+b^2+c^2+d^2)$

$=(x^2+y^2)(a^2+b^2)+(z^2+w^2)(c^2+d^2)+(x^2+y^2)(c^2+d^2)+(z^2+w^2)(a^2+b^2)$

$=(xa-yb)^2+(xb+ya)^2+(zc-wd)^2+(zd+wc)^2+(xc-yd)^2+(xd+yc)^2+(za-wb)^2+(zb+wa)^2$

$=[(xa-yb)^2)+(zd+wc)^2]+[(za-wb)^2+(xd+yc)^2]+[(xb+ya)^2+(zc-wd)^2]+[(xc-yd)^2+(zb+wa)^2]$

$=(xa-yb+zd+wc)^2+(wb-za+xd+yc)^2+(xb+ya+zc-wd)^2+(yd-xc+zb+wa)^2$

(I did skip a few steps because its very long)

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@grayguy I think you wanted something like this, if not then please do reply and i will delete it. – Abhra Abir Kundu Jun 11 '13 at 19:18
The above works for the four variable case, and looks to be quaternion multiplication as mentioned by Will. I think the OP wants info about the five variable case. I'm giving an up since this approach at least shows an understanding of what OP originally wanted. – coffeemath Jun 12 '13 at 0:50
@coffeemath At first there was four variable.....then op changed it to five variable.... – Abhra Abir Kundu Jun 12 '13 at 8:42
It seems to me this post has changed too much. This makes your answer above appropriate for the 4 variable case. – coffeemath Jun 12 '13 at 13:18