Is this product of 8 quaternions a real value?

Question

I'd like to know the product of 8 quaternions $$(a+bi+cj+dk)(a-bi+cj+dk)(a+bi-cj+dk)(a+bi+cj-dk)(a+bi-cj-dk)(a-bi-cj+dk)(a-bi+cj-dk)(a-bi-cj-dk)$$

I'd like to know if it is real?

PS : $a,b,c,d \in \mathbb{R}$

Answer 1

No. Let $a = 1, b = 2, c = 3, d = 4$, then your equation evaluates to $$550800+331200i+57600j-489600k$$

Also, notice that your equation has a slight asymmetry. You can see it in the signs: $$+++, -++, +-+, ++-$$ $$+--, --+, -+-, ---$$ notice how the 6th and 7th sign pattern seems to be swapped.

If you use this sign pattern, which is more symmetric: $$+++, -++, +-+, ++-$$ $$+--, -+-, --+, ---$$ then you'll get $195600+ 550400i -396800j+ 396800k$

In any case, the result is not real.

If you use this sign pattern: $$+++, -++, +-+, ++-$$ $$--+, -+-, +--, ---$$ then the quaternions do pair up into conjugates, without ever needing to swap places (swapping places is not usually allowed, since quaternions are not generally commutative), like this: $$q_{+++}q_{-++}q_{+-+}(q_{++-}q_{--+})q_{-+-}q_{+--}q_{---}$$ $$=q_{+++}q_{-++}q_{+-+}(a^2+b^2+c^2+d^2)q_{-+-}q_{+--}q_{---}$$ (since real numbers commutes with quaternions,) $$=(a^2+b^2+c^2+d^2)q_{+++}q_{-++}(q_{+-+}q_{-+-})q_{+--}q_{---}$$ $$=(a^2+b^2+c^2+d^2)^2q_{+++}(q_{-++}q_{+--})q_{---}$$ $$=(a^2+b^2+c^2+d^2)^3q_{+++}q_{---}$$ $$=(a^2+b^2+c^2+d^2)^4$$ which is real.