# Why does $ab=ba=1$ imply ${a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 = 1$?

Let's say that I've got a group $V$ of integer quaternions of the form $\mathbb{Z} + \mathbb{Z}i + \mathbb{Z}j + \mathbb{Z}k$. Now assume that there exists an element $a = a_1 + a_2i + a_3j + a_4k$ such that $ab=ba=1$ for some other element $b \in V$. Note also that Why would this imply that ${a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 = 1$?

• This is a consequence of the fact that the usual norm $||\,\cdot\,||$ on $\Bbb H$ is compatible with quaternionic multiplication in the sense that $||ab|| = ||a||\,||b||$. Commented Feb 1, 2016 at 1:36
• Yes I can see that and understand it but how do i use that fact to show this? Commented Feb 1, 2016 at 1:37
• I just don't see how this follows from that fact? Commented Feb 1, 2016 at 1:37
• Combine this fact with the observation that for any nonzero $a \in V$ we have $||a|| \geq 1$. Commented Feb 1, 2016 at 1:38
• In fact, it seems that $ab = 1$ is enough, so we can drop the additional hypothesis that $ab = ba$ (which gives a strictly stronger condition). Commented Feb 1, 2016 at 1:39

As Travis said in the comments,

$$\| a b\| = \| a\| \| b\|$$,

in this case we have $ab=1$ so we can write,

$$1 = \|a\| \|b\|$$,

and if we square both sides we can write,

$$1 = (a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2).$$

All the $a_n$ and $b_n$ are integers. We then have that the product of two integers is equal to $1$. Under what circumstances is this possible?