Find the inverse of all $\{1, 3, 5, 7\}$ modulo $8$.

Question

Find the inverse of all $\{1, 3, 5, 7\}$ modulo $8.$ What do you observe? Can you explain this?

I noticed that they are all inverses of themselves. But I do not know why this occur. Any help is appreciated.

Answer 1

Consider an odd number $x$ of the form $2n+1$. Then, $$x^2 - 1 = (2n+1)^2 - 1 = 4n^2 + 4n = 4n(n+1)$$ Since $2$ divides $n$ or $n+1$, we have $2 \mid n(n+1)$. As a result, $8 \mid 4n(n+1)$. Therefore, if $x$ is an odd natural number, $x^2 - 1$ is a multiple of $8$. Does this answer your question?