**An Infinite Family of ****Four-Square**** and ****Eight-Square**** Identities**

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**By Tito Piezas III**

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*Abstract*: An infinite family of 8-square identities (which can be specialized into a 4-square) will be given.

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**I. Introduction**

**II. A Family of 4-Square Identities**

**III. A Family of 8-Square Identities**

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**I. Introduction**

In the previous article, *Pfister’s 16-Square Identity*, aside from giving a non-bilinear 16-square identity, a variation of *Euler’s Four-Square Identity* with only two *bilinear* z_{i} was also mentioned, namely,

One can ask a reasonable question: *How many four-square identities are there?* It can be shown that there are an *infinite* number of four-square and eight-square identities where not all z_{i} are bilinear.

**II. A Family of 4-Square Identities**

In contrast to the Pfister 4-square, the one below generally has only *one* bilinear z_{i},

*for arbitrary n*. Of course, for *n* = *x*_{4}, then *k* = 0, all the z_{i} become bilinear, hence the *Euler 4-square* can be considered a special case of this family. The {a_{i}, b_{i}, c_{i}} also satisfy the nice relations,

a_{1}^{2} + b_{1}^{2} + c_{1}^{2} = x_{1}^{2} + x_{2}^{2} + x_{4}^{2}

a_{2}^{2} + b_{2}^{2} + c_{2}^{2} = x_{1}^{2} + x_{3}^{2} + x_{4}^{2}

a_{3}^{2} + b_{3}^{2} + c_{3}^{2} = x_{2}^{2} + x_{3}^{2} + x_{4}^{2}

and,

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = x_{2}^{2} + x_{3}^{2} + x_{4}^{2}

b_{1}^{2} + b_{2}^{2} + b_{3}^{2} = x_{1}^{2} + x_{3}^{2} + x_{4}^{2}

c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = x_{1}^{2} + x_{2}^{2} + x_{4}^{2}

If we include the leading coefficients of the first three z_{i} as {a_{0}, b_{0}, c_{0}} = {x_{1}, x_{2}, x_{3}}, then the last three relations simplify as,

**Σ ****a**_{i}^{2}** = Σ ****b**_{i}^{2}** = Σ ****c**_{i}^{2}** = Σ ****x**_{j}^{2}

However, this 4-square family in turn is just a *special* case of an 8-square family.

**III. A Family of 8-Square Identities**

An 8-square identity with four bilinear z_{i} was given in *Pfister’s 16-Square Identity*. But we can also give one with *five* bilinear z_{i}. First, define the following variables,

where,

*m*** = ***x*_{4}-*n*

and the other variables to be defined later. Also,

For the latter set, note that, for *any value* of the variables, then,

hence, they are just three versions of the *Euler 4-Square*. And where do the {*a*_{i}, *b*_{i}, *c*_{i}} appear? *In an 8-square identity*, namely,

**and we can now define {***p*_{i}, *q*_{i}, *r*_{i}, *s*_{i}, *t*} in terms of the *x*_{i} and *n* as,

*for t = ***0***, and arbitrary n*. (The variable *t* was just a place-holder.) With these definitions, it can be seen that,

and the {a_{i}, b_{i}, c_{i}} will also satisfy the additional nice relations,

a_{1}^{2} + b_{1}^{2} + c_{1}^{2} = x_{1}^{2} + x_{2}^{2} + x_{4}^{2}

a_{2}^{2} + b_{2}^{2} + c_{2}^{2} = x_{1}^{2} + x_{3}^{2} + x_{4}^{2}

a_{3}^{2} + b_{3}^{2} + c_{3}^{2} = x_{2}^{2} + x_{3}^{2} + x_{4}^{2}

a_{4}^{2} + b_{4}^{2} + c_{4}^{2} = x_{5}^{2} + x_{6}^{2} + x_{7}^{2}

a_{5}^{2} + b_{5}^{2} + c_{5}^{2} = x_{5}^{2} + x_{6}^{2} + x_{8}^{2}

a_{6}^{2} + b_{6}^{2} + c_{6}^{2} = x_{5}^{2} + x_{7}^{2} + x_{8}^{2}

a_{7}^{2} + b_{7}^{2} + c_{7}^{2} = x_{6}^{2} + x_{7}^{2} + x_{8}^{2}

as well as,

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} + a_{5}^{2} + a_{6}^{2} + a_{7}^{2} = x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2} + x_{6}^{2} + x_{7}^{2} + x_{8}^{2}

b_{1}^{2} + b_{2}^{2} + b_{3}^{2} + b_{4}^{2} + b_{5}^{2} + b_{6}^{2} + b_{7}^{2} = x_{1}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2} + x_{6}^{2} + x_{7}^{2} + x_{8}^{2}

c_{1}^{2} + c_{2}^{2} + c_{3}^{2} + c_{4}^{2} + c_{5}^{2} + c_{6}^{2} + c_{7}^{2} = x_{1}^{2} + x_{2}^{2} + x_{4}^{2} + x_{5}^{2} + x_{6}^{2} + x_{7}^{2} + x_{8}^{2}

Again, if we include the leading coefficients as {a_{0}, b_{0}, c_{0}} = {x_{1}, x_{2}, x_{3}}, then the last three relations simplify as,

**Σ ****a**_{i}^{2}** = Σ ****b**_{i}^{2}** = Σ ****c**_{i}^{2}** = Σ ****x**_{k}^{2}

While *n* is arbitrary, there is no value *n* such that all the z_{i} become bilinear, hence this 8-square family cannot reduce to the *Degen-Graves 8-square*. However, if we set {x_{i}, y_{i}} = 0 for all *i *> 4, then it does reduce to the 4-square family in the previous section where, as was pointed out already, the special case *n* = *x*_{4} further reduces it to the *Euler 4-square*. There is in fact an 8-square family that can reduce to the Degen-Graves, but its variables are too tediously complicated for this article. The one given suffices to prove there is an infinite number of 8-square identities.

**-- End --**

© Tito Piezas III, Feb 2012

You can email author at *tpiezas@gmail.com*.