Finding $7$ inverse modulo $11$ I'm trying to find the inverse of $7$ modulo $11$. From what I understand, the steps are: 

\begin{align}
&amp;11 = 1(7) + 3 \\
&amp;7 = 2(3) + 1 \\
\end{align}

From here, you work backwards

\begin{align}
1 &amp;= 7 - 2(3) \\
&amp;= 7 - 2(11-7) \\
&amp;= 7 - 2(11) + 2(7) \\
&amp;= -2(11) + 3(7)
\end{align}

Now, here I see that $7$'s inverse modulo is $3$ but there can be many more (right?). The answer book says that one of them is $8$. Now, assuming my previous steps are correct; how do I find the others? 

Thanks!

EDIT: 

\begin{align}
11 &amp;= 1(7) + 4 \\
7 &amp;= 1(4) + 3 \\
4 &amp;= 3(1) + 1
\end{align}

Then,

\begin{align}
1 &amp;= 4 - 1(3) \\
&amp;= 4 - 1(7 - 4) \\
&amp;= 4 - 7 + 4 \\
&amp;= -7 + 2(4) \\
&amp;= -7 + 2(11 - 7)\\
&amp;= -7 + 2(11) - 2(7) \\
&amp;= 2(11) - 3(7)
\end{align}
Hint $\ {\rm mod}\ 11\!:\ 7\equiv -\color{#c00}4,\,$ and $\ {\rm mod}\ 2n\!-\!1\!:\ 2n\equiv 1\,\Rightarrow\, 4n^2\equiv 1\,\Rightarrow\,\color{#c00}4^{-1}\equiv n^2$
$8 \times 7 = 56$

$56 \ \mbox{mod $7$} = 1$

Done.

$7^{-1} \ \mbox{mod $11$} = 8$