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"Prove that if $n$ is coprime to $10$ then $n^{101} \equiv n \pmod{1000}$"

I know that this has something to do with Euler's function, but i'm not sure how to apply it.

A fellow on an IRC channel that I am on answered this question, but I did not understand his process. Written below is his process.

"$n$ is prime to $10$ => $n^{\phi(125)+1} = n^{101} = n \pmod{125}$

since $n^2 = 1 \pmod{8}, n^{50*2+1} = n^{101} = n \pmod{8}$ => $n^{101} = n\pmod{125 * 8}$; Thus if $gcd(n,10)=1$ then $n^{101} = n \pmod{1000}$"

If $n$ is coprime to $10$, why are we allowed to say $n^{\phi(125)+1}$? Eueler's function says if $n$ and $a$ are coprime, then: $a^{\phi(n)} \equiv 1 \pmod{n}$.

I think I just need this part explained and i'll be okay.

Anirban Niloy
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Ozera
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2 Answers2

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Euler's theorem tells us $n^{\varphi(125)}\equiv 1\pmod{125}$. Multiplying both sides by $n$ yields $n^{\varphi(125)+1}\equiv n\pmod{125}$. As you can tell, $\varphi(125)=\varphi(5^3)=5^2(5-1)=100$. So far we have:$$n^{100+1}\equiv n\pmod{125}\\n^{101}\equiv n\pmod{125}$$Can you finish the rest? Because $n$ is coprime to $10$ it follows that it is also coprime to $2,5$ and also any of their powers -- which is how we knew $n$ and $5^3=125$ were coprime. You can use a similar argument for $n$ and $2^3=8$.

obataku
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  • Are we allowed to say that because $n$ is coprime with 5 and 2 as well as their powers, we can multiply $125$ in $n\pmod{125}$ by $2^3$ to get $n\pmod{125*8}$ => $n\pmod{1000}$. Is that correct reasoning? – Ozera Aug 31 '13 at 20:07
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    Right; if $a\equiv b\pmod{k}$ and $a\equiv b \pmod{l}$ and $gcd(k,l)=1$, then $a\equiv b\pmod{kl}$. – user84413 Aug 31 '13 at 20:53
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    @Ozera see the above comment. – obataku Aug 31 '13 at 20:55
  • @oldrinb Ah yes thank you. He just said what I said, but in a cleaner way. Thank you all. – Ozera Aug 31 '13 at 21:25
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We can use Carmichael function,

$\lambda(1000)=$lcm$(\lambda(5^3),\lambda(2^3))=$lcm $(100,2)=100$

Alternatively, as $n^{100}\equiv1\pmod{125}$

and as $n$ is odd, $n^2=(2m+1)^2=8\frac{m(m+1)}2+1\equiv1\pmod 8$

$\implies n^{lcm(100,2)}\equiv1\pmod {125}$ and $n^{lcm(100,2)}\equiv1\pmod 8$

$\implies n^{100}\equiv1\pmod {125}$ and $n^{100}\equiv1\pmod 8$

$\implies n^{100}-1 $ will be divisible by lcm$(125,8)=1000$

This is what is hidden in Carmichael function

lab bhattacharjee
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