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

Question

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 ${a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 = 1$?

Answer 1

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?