Can we find m in this modular?
$$aa^{-1} \equiv 1\mod\ m $$
$$where\ a ,a^{-1} are\ known$$
Asked
Active
Viewed 51 times
0
-
1Welcome to Math SX! $m$ is divisor of $aa^{-1}-1$, that's all we can say. – Bernard Oct 29 '17 at 01:28
-
1No, because the statement is true for all non-zero integers m – Francisco José Letterio Oct 29 '17 at 01:28
1 Answers
1
Suppose integer values are specified for $a,a^{-1}$.
To be less confusing, let $b = a^{-1}$.
Then $$ab\equiv 1\;(\text{mod}\;m) \iff\ m\mid (ab-1)$$
so any integer divisor of $ab-1$ qualifies as a value of $m$.
For example, let $a=7,\;a^{-1}=4$.
\begin{align*} \text{Then}\;\;&7^{-1}\equiv 4\;(\text{mod}\;m)\\[4pt] \iff\;&28 \equiv 1\;(\text{mod}\;m)\\[4pt] \iff\;&m\mid 27\\[4pt] \iff\;&m\in\{\pm 1,\pm 3,\pm 9,\pm 27\}\\[4pt] \end{align*}
quasi
- 58,772
-
-
Right. There are choices. Any integer divisor of $ab-1$ will qualify as a value of $m$. – quasi Oct 29 '17 at 01:44
-