Product as linear combination of squared linear combinations

Question

Edit:

This question is an old version and could be closed. More details can be found in this post.

Motivation

A product of two factors can also be written as linear combination of squared linear combinations

$$x_1x_2=\frac{1}{4}(x_1+x_2)^2-\frac{1}{4}(x_1-x_2)^2=\sum_{i=1}^{n_2} a_i \left(\sum_{j=1}^{2}b_{i,j}x_j\right)^2$$ with $n_2=2,a_1=\frac{1}{4},a_2=-\frac{1}{4},b_{1,1}=1,b_{1,2}=1,b_{2,1}=1,b_{2,2}=-1$.

Similarly, for 3 factors it was found by user Essaidi $$x_1x_2x_3=\frac{1}{24}(x_1+x_2+x_3)^3−\frac{1}{24}(−x_1+x_2+x_3)^3−\frac{1}{24}(x_1−x_2+x_3)^3−\frac{1}{24}(x_1+x_2−x_3)^3$$

However this simple pattern does not continue.

Question

Which real coefficients $a_i,b_{i,j}$ fulfill the product of four factors? $$x_1x_2x_3x_4=\sum_{i=1}^{n_4} a_i\left( \sum_{j=1}^{4}b_{i,j}x_j \right)^4$$

General Question

Which real coefficients $a_i,b_{i,j}$ fulfill the product of $k$ factors with $k\in \mathbb{N}^+$ ?

$$\prod_{j=1}^k x_j=\sum_{i=1}^{n_k} a_i\left( \sum_{j=1}^{k}b_{i,j}x_j \right)^k$$

Note:

The smallest possible $n_k$ is of interest. It is allowed for some $b_{i,j}=0$.

It is a bad habit to alert the question after others have posted an answer. Please consider asking another question for the case $k\ge 4$ and restore the post to the original question. — Arctic Char, Dec 31 '21 at 23:13
I have reverted this question to a point where the question and answer now align. If you have a different question to ask, I suggest writing a different answer. — davidlowryduda, Jan 04 '22 at 18:26
The title and the text of the question do not fit. The title and first sentence speak about squares, but in the rest of the text we find cubics and higher powers. I will place a new question but fear that this might be not possible due to similarity. — granular_bastard, Jan 05 '22 at 09:50

score 3 · Answer 1 · answered Dec 31 '21 at 20:06

3

There aren't such coefficients when $k \geq 3$. In fact consider the case : $$x_1 = x_2 = \cdots = x_k = x$$ We will have : $$x^k = \sum_{i = 1}^{n_k} a_i \left(\sum_{j = 1}^k b_{i, j}\right)^2 x^2$$ Contradiction, because we have two polynomials with different degrees.
It only works when $k = 2$.

answered Dec 31 '21 at 20:06

Essaidi

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Suppose the formula works and apply it to $\lambda x_1, \ldots, \lambda x_k$. You will have $\lambda^k$ in the LHS and only $\lambda^2$ in the RHS. Necessarily k must be 2 – Essaidi Dec 31 '21 at 20:52
In case $k = 3$ we have : $$x y z = \dfrac{1}{24} \left((x + y + z)^3 - (-x + y + z)^3 - (x - y + z)^3 - (x + y - z)^3\right)$$ – Essaidi Dec 31 '21 at 21:20

Product as linear combination of squared linear combinations

1 Answers1