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I want to find a solution $x \in \mathbb{N}$ of $x^{125} \equiv 7 \ (\mathrm{mod} \ 39)$ by using Ender Wiggin's method Find $x^{98}$ congruent to $7$ (mod $18$)

My steps are:

$x^{125} \equiv 7 \ (\mathrm{mod} \ 39) \Rightarrow x^{125}=7+39 \cdot n,\ n \in \mathbb{N}$

$\Rightarrow \mathrm{gcd}(x,39)=1$

It's $\varphi(39)=\varphi(3)\cdot\varphi(13)=24$

So $x^{24} \equiv 1 \ (\mathrm{mod} \ 39)$

Since $x^{125} \equiv 7 \ (\mathrm{mod} \ 3)$ and $x^{125} \equiv 7 \ (\mathrm{mod} \ 13)$ it follows that

$x^{\varphi(3)}=x^2 \equiv 1 \ (\mathrm{mod} \ 3)$ and $x^{\varphi(13)}=x^{12} \equiv 1 \ (\mathrm{mod} \ 13)$.

Now I don't know what to do next.

I tried with $x=4$ and it works, but how to show it properly?

What has to be done now?

Tartulop
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  • $39=3\times 13$ so it suffices to solve the problem $\pmod 3$ and $\pmod {13}$. – lulu May 16 '20 at 13:31
  • By modular order reduction it reduces to $,x^5\equiv 7,$ by $,5 = 125\bmod 24,,$ by $, 24=\phi(39).,$ As in the dupe, we can take the $5$'th root by raising to power $,1/5\equiv 5/25\equiv 5\pmod{!24},$ so $,x\equiv 7^5\equiv -2\pmod{!39}$ – Bill Dubuque May 16 '20 at 16:08

2 Answers2

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Better use Carmichael Function $$\lambda(39)=12$$

and $125\equiv5\pmod{12}$

$$ x^{125}\equiv x^5\pmod{39}$$

$$\implies x^5\equiv7\pmod{39}$$

Now as $x^{12}\equiv1\pmod{39}$

$$x\equiv(x^5)^5(x^{12})^{-2}\equiv7^5\cdot1^{-2}\pmod{39}$$

Finally $7^2\equiv10,7^4\equiv10^2\equiv-17, 7^5\equiv7(-17)\equiv-2\equiv39-2$

lab bhattacharjee
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    Carmichael doesn't gain anything, by $,1/5\equiv 25/5\equiv 5,$ is the same mod $24=\phi(n)$ as it is mod $12=\lambda(n)\ \ $ – Bill Dubuque May 16 '20 at 16:36
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$ x^{24} \equiv 1 \pmod{39} $ is not something to be solved; it's something you know is true whenever $\gcd(x,39)=1$. So

$$ x^{125} = (x^{24})^5 x^5 \equiv x^5 \equiv 7 \pmod{39} $$

Then you can continue as in @lab bhattacharjee's answer.

Or for a different approach, you could start by noting that $x^{125} \equiv 7 \pmod{39}$ if and only if $x^{125} \equiv 7 \pmod{3}$ and $x^{125} \equiv 7 \pmod{13}$.

aschepler
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  • Since $x^{125} \equiv 7 \ (\mathrm{mod} \ 3)$ and $x^{125} \equiv 7 \ (\mathrm{mod} \ 13)$ it follows that $x^{\varphi(3)}=x^2 \equiv 1 \ (\mathrm{mod} \ 3)$ and $x^{\varphi(13)}=x^{12} \equiv 1 \ (\mathrm{mod} \ 13)$. Now I don't know what to do next. I tried with $x=4$: $ 16 \equiv 1 \ (\mathrm{mod} \ 3)$ and $ 16777216 \equiv 1 \ (\mathrm{mod} \ 13)$, so it works. But how to show it properly? – Tartulop May 16 '20 at 14:21
  • Use the same idea of reducing the power. Since $x^2 \equiv 1 \pmod 3$, $7 \equiv x^{125} = x^{124} x = (x^{2})^{62} x \equiv 1^{62} x = x \pmod 3$. Since $x^{12} \equiv 1 \pmod{13}$, $7 \equiv x^{125} = x^{120} x^5 = (x^{12})^{10} x^5 \equiv 1^{10} x = x \pmod{13}$. – aschepler May 16 '20 at 14:35
  • 4 "works" for those two equations just because 4 is relatively prime to both 3 and 13. They will also be true for 2, 5, 7, 8, 10, .... but you're just verifying Fermat's Little Theorem. Again, the equations coming from Euler's Theorem aren't the things you want to solve; they're the helpful facts you want to use. – aschepler May 16 '20 at 14:39
  • So $x^{125} \equiv x \equiv 7 \ (\mathrm{mod} \ 3)$ and $x^{125} \equiv x \equiv 7 \ (\mathrm{mod} \ 13)$. But how to conclude that $x=-2$, such that $-2^{125} \equiv 7 \ (\mathrm{mod} \ 39)$? – Tartulop May 17 '20 at 15:38