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I found this question in my book but got no idea on how to approach it.

  1. Find one x that solves: $x^7=4(mod 81)$
  2. How many different x's between 1 and 80 solve the equation?

I know that $x^{\Phi(81)}=x^{54}=1 (mod 81)$ but that doesn't help. Plus I tried calculating for small number like 1,2,3 yet nothing worked...

Note: if 54 was multiplication of 7 then I could solve it easily as I know that $x^{54}=x^{7k}={x^{k}}^{7}$ so the final answer would be $4*x^k (mod 81)$

My answer to 1:

I know that $2^54 = 1 (mod 81)$ so $2^56 = 4 (mod 81)$ which means: ${2^{8}}^{7} = 4 (mod 81)$ and the answer is $2^8 (mod 81)$ which is equal to 13.

But what about 2? I have no idea at all...

Bill Dubuque
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Algo
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    Since $81 = 3^4$, you can reduce the equation modulo $3$ to get $x^7 \equiv 1 \pmod 3$, so $x \equiv 1 \pmod 3$, hence $x$ has the form $3 y + 1$, cutting your search space by $2 / 3$. Now, substitute $x = 3 y + 1$ in the original equation and reduce modulo $9$ to produce a restriction on $y$, etc. – Travis Willse Jul 30 '23 at 22:33
  • @Algo FYI, there's what I consider to be a simpler approach for your particular problem. Note an Approach0 search comes up with how to solve this kind of modular arithmetic problems with exponents?, where you can use a similar approach to what's in its answer. First, though, note that, using Euler's totient function, $\varphi(81)=81\left(1-\frac{1}{3}\right)=54$, not $78$ as you indicate. ... – John Omielan Jul 30 '23 at 22:46
  • (cont.) Thus, using $4=2^2$, we can get by taking both sides to the power $27$ that $x^{27}\equiv 1\pmod{81}$, then we can take both sides of the original congruence equation to the power of $4$ to finish off. – John Omielan Jul 30 '23 at 22:47
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    Since $\gcd(7, 54)=1$, there can't be a $7^{th}$ root of unity $\pmod {81}$, so every residue prime to $81$ has a unique $7^{th}$ root. – lulu Jul 30 '23 at 22:59
  • @JohnOmielan Thanks that did help a lot, please see my answer and let me know if it's correct now :), can you please help with 2 how can I count all possible answers without testing every value? – Algo Jul 30 '23 at 22:59
  • @Algo You're welcome. You've correctly determined $x\equiv 13\pmod{81}$ is one solution. Note lulu's comment, and also cubesteak's answer, both indicate how to show there's just one solution. Alternatively, my suggestion, from my earlier comment, is that $x^{28}\equiv 4^4\equiv 13\pmod{81}$, so $x^{27}\equiv 1\pmod{81}$ means $x\equiv 13\pmod{81}$, with this method explicitly showing there is just one solution. In either case, this means you can then answer your question $2$. – John Omielan Jul 30 '23 at 23:06
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    Just as a matter of notation, exponentiation associates on the right. That is, when you write ${2^{8}}^{7}$, it means ${2^{(8^7)}}$, not $({2^{8}})^{7}$. – David K Jul 30 '23 at 23:15
  • See the Theorem in the linked dupe. – Bill Dubuque Jul 31 '23 at 01:13

1 Answers1

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You know that $x^{54}\equiv1\pmod{81}$. What if we use that by exponentiating both sides of the equivalence? First, solve $7k\equiv1\pmod{54}$. We get $k\equiv1\pmod2$ and $k\equiv4\pmod{27}$, so $k\equiv31\pmod{54}$. So $$x^{31(7)}\equiv4^{31}\pmod{81}$$ $$x^{4(54)+1}\equiv2^{62}\pmod{81}$$ $$x\equiv2^8\pmod{81}$$

Mike
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