Can the expression $a^2 + b^2 - c^2$ be factored as a product of two quaternions, where $a$, $b$, $c$ are real numbers?

Question

I'm trying to factor the expression $$a^2+b^2-c^2$$ as a product of two quaternions, where $a,b,c$ are reals. Can anyone give me the answer?

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I think it can't be done. I started with multiplying $$(Aa,Bb,Cc,Dd)(Ea,Fb,Gc,Hd)=\\ (a^2 A E - b^2 B F - c^2 C G - d^2 D H,\\ a b B E + a b B F - c d D G + c d C H, \\ a c C E + b d D F + a c A G - b d B H, \\ a d D E - b c C F + b c B G + a d A H)$$ For some real numbers $A, B, \dotsc, H$ , where I use the convention $(A,B,C,D)=A +Bi +Cj+Dk$. Since we want the result to be real for all $a,b,c,d$ we can write $$D E + A H = 0\\ C F - B G = 0\\ C E + A G =0\\ D F -B H = 0\\ B E + A F = 0\\ D G - C H = 0$$ What remains is to set the real part to be equal to the desired result. For example if we choose the polynomial $a^2+b^2+c^2$ we get $$a^2+b^2+c^2=(0,a,b,c)(0,-a,-b,-c).$$ However, my question was to find the factorization for $$a^2+b^2-c^2.$$ In that case the system doesn't have any real solutions for $A,B,\dotsc,H$.

I suspect therefore that all the polynomials, where one sign differs from the others, e.g. $$a^2-b^2+d^2\\ b^2+c^2-d^2$$ will also not be factorisable. Can anyone confirm this?

Answer 1

It cannot be written as the product of two $\mathbb{H}$-linear combinations of variables $a,b,c$. (We assume the formal variables commute with all quaternion scalars, of course.)

Suppose it were, say $(ap+q)(ar+s)$ where $p,q\in\mathbb{H}$ and $r,s$ are $\mathbb{H}$-linear combinations of $b$ and $c$. The leading term (with respect to $a$) would then be $pra^2$ which must be $a^2$ and so $pr=1$. We may factor the coefficients of $a$ out as $(a+qp^{-1})pr(a+r^{-1}s)=(a+u)(a+v)$. Now the middle term is $(u+v)a$, which must be $0$ as in the expression $a^2+b^2-c^2$, so $v=-u$ and we get $(a+u)(a-u)=a^2-u^2$. Thus, we want some combination of $b$ and $c$ (that is, $u$) which squares to $b^2-c^2$.

Writing $(xb+yc)^2=b^2-c^2$, we see $x^2=1$ so $x=\pm1$. But then the middle term is $\pm2ybc$ which can only be $0$ if $y=0$ in which case we don't see $-c^2$ as the final term.

It is however factorizable in the split quaternions. Then $a^2+b^2-c^2-d^2$ is the product of the split quaternion $a+bi+cj+dk$ with its conjugate $a-bi-cj-dk$.