Solving $217 x \equiv 1 \quad \text{(mod 221)}$

Question

I am given the problem:

Find an integer $x$ between $0$ and $221$ such that $$217 x \equiv 1 \quad \text{(mod 221)}$$

How do I solve this? Unfortunately I am lost.

Look up the Extended Euclidean Algorithm and Bezout's Identity. They are useful for solving this type of problem. — robjohn, Mar 30 '12 at 17:22

score 8 · Accepted Answer · edited Apr 13 '17 at 12:19

8

In this special case, you can multiply the congruence by $-1$ and you'll get $$4x\equiv 220 \pmod{221}.$$ (Just notice that $-217 \equiv 4 \pmod{221}$ and $-1\equiv220\pmod{221}$.)

This implies that $x\equiv 55 \pmod{221}$ is a solution. (And since $\gcd(4,221)=1$, there is only one solution modulo $221$.)

In general, for questions of this type you can use extended Euclidean algorithm see Wikipedia.

You can find some examples at this site, e.g. here.

edited Apr 13 '17 at 12:19

Community

1

answered Mar 30 '12 at 15:15

Martin Sleziak

53,687

can you elaborate on we get from multiplying by -1, to $4x \equiv 220 \text{(mod 221)}$? I see the parenthesis, but can say that more explicitly? having a little trouble following – Dominick Gerard Mar 30 '12 at 15:20
$-217$ and $4$ are congruent modulo $221$, hence we can replace $-217$ by $4$ without changing the validity of the congruence relation. The same is true for $-1$ and $220$. – Martin Sleziak Mar 30 '12 at 15:23
Although in this case we were able to make a guess which made finding the solution easy, it is useful to learn a method how to do this in general. – Martin Sleziak Mar 30 '12 at 15:32

score 3 · Answer 2 · answered Mar 30 '12 at 17:13

Using the Euclid-Wallis Algorithm: $$ \begin{array}{r} &&1&54&4\\ \hline \color{red}{1}&0&1&\color{red}{-54}&217\\ 0&\color{green}{1}&-1&\color{green}{55}&-221\\ \color{red}{221}&\color{green}{217}&4&\color{blue}{1}&0 \end{array} $$ we get that $\color{green}{55\cdot217}\color{red}{-54\cdot221}=\color{blue}{1}$. Thus, $x=55\pmod{221}$.

Bill Dubuque · Answer 3 · 2016-12-12T02:36:10.880

Hint $\ {\rm\ mod}\,\ 221\!:\quad\ \dfrac{1}{217}\equiv\dfrac{-220}{-4}\equiv 55$

${\rm mod}\,\ 4n\!+\!1\!:\ \ \dfrac{1}{4n\!+\!1\!-\!4}\equiv \dfrac{-4n}{-4}\equiv n\quad $ [above is case $\,n = 55$]

${\rm mod}\,\ an\!+\!1\!:\,\ \dfrac{1}{an\!+\!1\!-\!a}\equiv \dfrac{-an}{-a}\equiv n\quad $ [above is case $\,a=4$]

i.e. $\ \ \ an\!+\!1\equiv 0\,\Rightarrow\, (-a)n\equiv 1\ \Rightarrow\ n \equiv \dfrac{1}{-a\ }\equiv \dfrac{1}{-a + k(an\!+\!1)}$

Solving $217 x \equiv 1 \quad \text{(mod 221)}$

3 Answers3