Find $x$ so $x^{31}\equiv 2 \pmod{81}$

Question

$81$ isn't a prime so I can't use Fermat's little theorem.

$81=3^{4} \Rightarrow \varphi(81)=54$ but about the $gcd(x,81)$ , I don't know if it's equals to $1$ or not

and euler theorem won't help because $x^{54}\equiv 1 \pmod{81}$ and I need to find $x^{31}$

How do I approach this problem?

Hint: 1) Find the modular inverse $m$ of $31$ modulo $54$. 2) Show that $x\mapsto x^{31}$ and $x\mapsto x^m$ are inverse of each other in $\Bbb{Z}_{81}^*$. — Jyrki Lahtonen, Oct 06 '16 at 15:38
See also: http://math.stackexchange.com/questions/1952272/how-to-solve-2n2-3n-1-equiv0-pmod250 — lab bhattacharjee, Oct 06 '16 at 15:44
Correct. Why does it then follow that if $x^{31}\equiv a\pmod{81}$ with $\gcd(3,a)=1$ then $a^7\equiv x\pmod{81}$? In other words, did you also do part 2. — Jyrki Lahtonen, Oct 06 '16 at 19:26

score 2 · Accepted Answer · answered Oct 06 '16 at 15:37

2

As $(31,54)=1$ we find by trial $1=31\cdot7-4\cdot54$

$$x^{31\cdot7-4\cdot54}=(x^{31})^7\cdot(x^{54})^{-4}\equiv2^7\cdot1^{4}\pmod{81}$$

answered Oct 06 '16 at 15:37

lab bhattacharjee

274,582

@JyrkiLahtonen, Are you talking about $7,-4$ then find my answer here : http://math.stackexchange.com/questions/407478/solving-a-linear-congruence – lab bhattacharjee Oct 06 '16 at 15:41
@JyrkiLahtonen, My answer is directed to all my readers as well :) I'm also ready to respond to the questions of my readers. – lab bhattacharjee Oct 06 '16 at 15:45
But you use euler theorem ,how can you make sure that $gcd(x,81)=1$? – Asaf Oct 06 '16 at 15:52
@Asaf, Good Question. If $3$ divides $x,3$ must divide $x^{31}$ and definitely $$3\nmid{x^4-2}$$ But we need $81|(x^{31}-2)\implies3$ must divide $$x^4-2$$ – lab bhattacharjee Oct 06 '16 at 15:56
$\require{cancel}$ If $3\mid x$ so $3^{31} \mid x^{31}$.
How and why you conclude that $3 \cancel{\mid} x^{4} - 2$ ?
– Asaf Oct 06 '16 at 16:16
@Asaf, As $2$ is not divisible by $3$ – lab bhattacharjee Oct 06 '16 at 16:23
I don't understand this: we need $81\mid (x^{31}-2)$ so 3 must divide $x^{4}-2$ $$$$ How do you get to $x^{4}-2$ ? – Asaf Oct 06 '16 at 16:41
@Asaf, Sorry for the typo. It should be $$x^{31}$$ – lab bhattacharjee Oct 06 '16 at 16:42
Let's see if I get it:
if $3 \mid x$ so $3^{31} \mid x^{31}$ $\Rightarrow 3^{4} \cdot 3^{27}\mid x^{31} \Rightarrow 3^{4} \mid x^{31}$

and we need $81\mid (x^{31}-2)$,as $2$ isn't divisible by $3$ so $81\cancel{\mid} (x^{31}-2)$ and there is a contradiction.

therefore $gcd(x,81)=1$
– Asaf Oct 06 '16 at 17:06
can you give me a confirmation? – Asaf Oct 06 '16 at 18:13
@Asaf, Confirmed – lab bhattacharjee Oct 07 '16 at 01:30

score 2 · Answer 2 · answered Oct 06 '16 at 17:21

This question is screaming for Hensel's Lemma. Let $f(x) = x^{31}-2$. Solve $f(x) \equiv 0 \pmod{3}$ and get $x\equiv 2 \pmod{3}$. Then lift to 9 to get $x\equiv 2 \pmod{9}$. Then lift to 27 to get $x\equiv 20 \pmod{27}.$ And finally lift to 81 to get $x\equiv 47 \pmod{81}$.

Find $x$ so $x^{31}\equiv 2 \pmod{81}$

2 Answers2