Solving the congruence $9x \equiv 3 \pmod{47}$

Question

For this question $9x \equiv 3 \pmod{47}$.

I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$ when $a=47$ and $b=9$

I subbed back into the given equation: \begin{align*} (9)(21) & \equiv 3 \pmod{47}\\ 189 & \equiv 3 \pmod{47}\\ 63 & \equiv 1 \pmod{47} \end{align*} and I'm stuck, should I divide $63$ by $9$ to get $7$? but it does not comply to the given equation as when $x=7$, it would become $16 \pmod{47}$.

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score 2 · Answer 1 · answered Mar 12 '18 at 13:44

2

you can write $$x\equiv \frac{3}{9}=\frac{1}{3}\equiv \frac{48}{3}\equiv 16\mod 47$$

answered Mar 12 '18 at 13:44

Dr. Sonnhard Graubner

95,283

José Carlos Santos · Answer 2 · 2018-03-12T14:57:45.193

1

You are overcomplicating this. First of all$$9x\equiv3\pmod{47}\iff47\mid3.(3x-1)\iff47\mid3x-1.$$Besides, the inverse of $3$ $\pmod{47}$ is$16$. So$$47\mid3x-1\iff47\mid16.(3x-1)\iff47|x-16.$$So, the solutions are thos integers $x$ such that $x\equiv16\pmod{47}$.

edited Mar 12 '18 at 14:57

answered Mar 12 '18 at 13:44

José Carlos Santos

427,504

score 1 · Answer 3 · answered Mar 12 '18 at 14:08

We wish to solve the congruence $9x \equiv 3 \pmod{47}$. You correctly found that $21$ is the multiplicative inverse of $9$ modulo $47$. However, you made an error in this step: $$9 \cdot 21 \equiv \color{red}{3} \pmod{47}$$ If we multiply one side of the congruence by $21$, we must multiply the other side of the congruence by $21$. You should have obtained \begin{align*} 21 \cdot 9x & \equiv 21 \cdot 3 \pmod{47}\\ 1x & \equiv 63 \pmod{47}\\ x & \equiv 16 \pmod{47} \end{align*}

score 1 · Answer 4 · answered Mar 12 '18 at 14:14

By Bezout's identity since $\gcd(9,47)=1$ we know that exist $a,b$ integers such that

$$9a+47b=1$$

then

$$\iff 9a=1-47 b \iff 9a\equiv 1 \pmod{47}$$

The integer "$a$" is defined the multiplicative inverse of $9 \mod 47$ and can be easily found by Euclidean Algorithm. In this case we found that

$21\cdot9-4\cdot 47=1\implies a=21$

Once we have the inverse we can find the solution that is

$9x \equiv 3 \pmod{47}\iff 21\cdot9x \equiv 21\cdot3 \pmod{47}\iff x\equiv 63\equiv 16 \pmod{47}$

score 1 · Answer 5 · answered Mar 12 '18 at 14:39

Using the Extended Euclidean Algorithm as implemented in this answer, $$ \begin{array}{r} &&5&4&2\\\hline 1&0&1&-4&9\\ 0&1&-5&21&-47\\ 47&9&2&1&0\\ \end{array} $$ shows that $21\cdot9-4\cdot47=1$ so we want $63\cdot9-12\cdot47=3$ which can be reduced, by subtracting $47$ from $63$ and $9$ from $12$, to $$ 16\cdot9-3\cdot47=3 $$

Solving the congruence $9x \equiv 3 \pmod{47}$

5 Answers5