Solving a congruence, tricky implication

Question

We want to prove that

$$243x \equiv 1 \mod 2018 \implies x^{403} \equiv 3 \mod 2018$$

My try :

Assume that $243x \equiv 1 \mod 2018$

We have $x^{2016} \equiv 1 \mod 2018$ (by Fermat ($1009$ is prime) and oddness of $x$) so $x^{2015} \equiv 243 \mod 2018$

but $403 \times 5 = 2015 $

hence $(x^{403})^5 \equiv 243 \mod 2018$ or equivalently $(x^{403})^5 \equiv 3^5 \mod 2018$. I wonder how to get from this last congruence to the desired $x^{403} \equiv 3 \mod 2018$ ?

Thanks for any suggestions.

Hint: can you do something with the very first equation you have once you notice that $243=3^5$? — Mark Bennet, May 20 '19 at 12:10
You are asking how to go from $(x^{403})^{5} \equiv 3^5 \mod 2018$ to $x^{403} \equiv 3 \mod 2018$, right? — acarturk, May 20 '19 at 12:18
Raise $\ 3^{\large 5} x \equiv 1\ $ to power $403,,$ then scale by $,3\ \ $ — Bill Dubuque, May 20 '19 at 13:31
Thanks everyone @MarkBennet It was easy based on your comment, but I was blind :-/ — ahmed, May 20 '19 at 13:47
Sorry, there was a typo in my answer ($x$ vs $x^k) $ which is now fixed. — Bill Dubuque, May 20 '19 at 13:56
I also added a remark on how it is related to computing modular $k$'th roots - which is very useful to know (follow the link). — Bill Dubuque, May 20 '19 at 14:12

score 2 · Answer 1 · answered May 20 '19 at 13:37

2

Hint:

$3^5x\equiv1 \implies x\equiv 3^{-5} \implies x^{403} \equiv 3^{-2015} $

and $3^{2016}\equiv1\pmod{2018}$

answered May 20 '19 at 13:37

J. W. Tanner

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score 1 · Accepted Answer · answered May 20 '19 at 13:28

From the last step of your attempt i.e. $(x^{403})^5 \equiv 3^5 \mod 2018$, you can go to the conclusion if you can show that $y^5 \equiv 243 \mod 2018$ has only one solution mod $2018$ (because this would force $x^{403} \equiv 3 \mod 2018$ as they are both solutions).

While that statement is true, its proof is not going to be easy because it boils down to asking why $x^5 \equiv 1 \mod 2018$ has a unique solution, and this is not clear at all.

Therefore, you have not done anything wrong but got yourself in a higher power than required.

However, you did do some groundwork. For example, by noting that $243x \equiv 1 \mod 2018$, so we may raise both sides to the power $403$ : $$ 3^5x \equiv 1 \mod 2018 \implies 3^{2015}x^{403} \equiv 1 \mod 2018 $$

Now, multiply by a further $3$, and eliminate the $3^{2016}$ using an argument made in your attempt.

Bill Dubuque · Answer 3 · 2019-05-20T14:13:43.493

It's a special case of the following

Lemma $ $ if $\,\gcd(a,n) = 1\,$ and $\, -\color{#c00}{jk}\equiv 1\pmod{\phi(n)}\,$ then $\ a^{\large j} x\equiv 1\,\Rightarrow\, x^{\large k}\equiv a\pmod{\!n}$

Proof $\ $ Raising $\, x\equiv a^{\large -j}\,$ to power $\,k\,$ yields $\,x^{\large k}\equiv a^{\large \color{#c00}{-jk}}\equiv a\,$ by $\, -\color{#c00}{jk}\equiv 1\pmod{\!\phi(n)}\ $ by modular exponent reduction.

Remark $ $ In fact $\, x^{\large k}\equiv a\iff x\equiv a^{\large 1/k}\equiv a^{\color{}{-j}}\ $ i.e. we can easily compute $k$'th roots when $k$ is coprime to $\,\phi(n)\,$ by raising to power $\,1/k,\,$ just like in $\Bbb R,\,$ see this Theorem.

Solving a congruence, tricky implication

3 Answers3