# Modular equations and Euler's theorem

Looking at group $$U{81}$$ find $$0 such that:

$$x^{19}=8 \:(mod\:81)$$

I know that $$o(8)=18$$, is it true to say:

$$(x^{19})^{18}=1 \:(mod\:81)$$

and from here to use Euler's theorem which yields:

$$x^{18}=1 \:(mod\: 81) \ ?$$

• Yes, it's true, but I'm not sure it's useful. Instead, I'd start with $x^{19}\equiv8\bmod3$ and work my way up to moduli $3^2,3^3,3^4$. "Hensel's Lemma" Jun 1 at 10:21
• I'm not familiar with that lemma, will check it out but why it doesn't useful? I already know a number that his 18 power is equivalent to 1 (8) Jun 1 at 10:25
• @3xhaust Hensel's lemma is a lemma that allows one to lift congruences $\pmod{p}$ to $\pmod{p^2}$. This allows us to work with smaller congruences which might be easier to work. I think your approach is also fine but it's a bit difficult to find all the elements of order 18 directly. Jun 1 at 11:00
• Not just to $p^2$, @dar, but to $p^3$, $p^4$, .... Jun 1 at 11:05
• Use $x\equiv y\pmod{n}$ for $x\equiv y\pmod{n}$. Jun 1 at 11:14

$$8$$ isn't the only integer with order $$18$$, so solving $$x^{18}\equiv 1\pmod{81}$$ gives solutions which don't solve $$x^{19}\equiv 8\pmod{81}$$.

You can note that $$(9-1)^{18}\equiv 1\pmod{81}$$ to find the answer as $$x=8$$.

• Cam you elaborate more please? I didn't understand how to use what you suggested Jun 1 at 11:57
• @3xhaust; as $8^{18}\equiv1\pmod{81}$ then $8^{19}\equiv8\pmod{81}$
– JMP
Jun 1 at 12:10
• But how can I find all the numbers that satisfy it? Jun 1 at 13:32
• @3xhaust; all numbers of the form $9k\pm1$.
– JMP
Jun 1 at 13:38
• What's the explanation for that? Especially for the +1? Jun 1 at 13:45