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

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? 
 [/INST] 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 $a-bi-cj-dk$.
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