Three squares

Proof 1 :

Since identities of the form

$\left( \sum x_i^2 \right) \left( \sum y_j^2 \right) = \sum z_j^2$

hinge on the fact that hypercomplex numbers (which form associative groups under multiplication) can only have $2^k$ basis vectors (that is to say, numbers like $1,i,j,k$ and so on), since $3 \neq 2^k$ we cannot have a group with $S$ .

Famous identities include those due to Fibonacci/Brahmagupta (2 squares), Euler (4), and Degen (8). They can be proved by algebraic manipulation, which is tedious, or by observing the product of two elements in $\mathbb{C},\mathbb{H},\mathbb{O}$ and so on, then taking the absolute value of the product.

Proof 2 :

As long as one pair $m,n \in S$ satisfies $mn \not \in S$ , the monoid structure is lost. We know that all positive integers can be expressed as the sum of three squares but $4^a(8b+7)$ . Clearly then, $m$ can be $2(8b+7)$ and $n = 2$ , so that the closure axiom is violated. Other constructions are possible.