We look for zero products of the form (A±B)(C±D) where A, B, C and D are all products of basis elements of V.
Letting a, b, c and d be the corresponding binary number and ignoring the sign or the term for now we see that the product will have 4 terms which must cancel by twos to produce 0.
This means that a⊕b⊕c⊕d = 0.
‘⊕’ is exclusive-or.
Our search logic has already produced the zero factors of the form x(A+B) which means that we will not find new values for a⊕b.
Ditto c⊕d.
This leaves 16 values (times two for sign ambiguity) for b that might lead to new 0 products.
<p>Searching this small space for CL(4) Clifford numbers that my inversion code cannot invert finds 10 values:
<br>1±γ<sub>i</sub>γ<sub>j</sub>γ<sub>k</sub> for i&lt;j&lt;k,
and 1±γ<sub>0</sub>γ<sub>1</sub>γ<sub>2</sub>γ<sub>3</sub>.
Generalizations to higher order Clifford spaces are unclear from this.
<p>Indeed for i&lt;j&lt;k (1+γ<sub>i</sub>γ<sub>j</sub>γ<sub>k</sub>)(1−γ<sub>i</sub>γ<sub>j</sub>γ<sub>k</sub>) = 0 = (1−γ<sub>i</sub>γ<sub>j</sub>γ<sub>k</sub>)(1+γ<sub>i</sub>γ<sub>j</sub>γ<sub>k</sub>).
This detour has not revealed any more zero divisors not already evident from symmetry.
It has ruled out a category of zero divisors.
It reminds us to include some we have omitted, however.
A superset of these, however, is the zero-divisors which are related to known divisors by orthogonal transformations on V.
<hr><pre>((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . 0) -1 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . -1) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . -1) (0 . 0) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . -1) ((0 . 0) 0 . 0) (0 . 0) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . 0) 0 . 1)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . 0) 1 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . 1) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 1) (0 . 0) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 1) ((0 . 0) 0 . 0) (0 . 0) 0 . 0)
((((1 . 0) 0 . 0) (0 . 0) 0 . 0) ((0 . 0) 0 . 0) (0 . 0) 0 . -1)</pre>