- #1

- 647

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First of all, notice that, by Euler's theorem, ##k^{4\cdot 5^{n-1}}\equiv1\pmod{5^n}## and ##k^{2^{n-1}}\equiv1\pmod{2^n}##. This makes it clear by induction that ##a\equiv b\pmod{4\cdot 5^{n-1}}\rightarrow k^a\equiv k^b\pmod{5^n}## and ##a\equiv b\pmod{2^{n-1}}\rightarrow k^a\equiv k^b\pmod{2^n}##.

Let ##n\ge2## and ##a\equiv b\pmod{10^n}##. Then, as ##\left.2^{n-1},4\cdot 5^{n-1}\right|10^n##, ##a\equiv b\pmod2^{n-1}## and ##a\equiv b\pmod5^{n-1}##, so ##k^a\equiv k^b\pmod{2^n}## and ##k^a\equiv k^b\pmod{5^n}##. Therefore ##k^a\equiv k^b\pmod{\mathrm{lcm}\left(2^n,5^n\right)}##, so ##k^a\equiv k^b\pmod{10^n}##.

Letting ##a=10^n## and ##b=0##, we get ##k^{10^n}\equiv k^0=1\pmod{10^n}##.

As this holds for all ##k## coprime to ##10^n##, this means ##\left.\lambda\left(10^n\right)\right|10^n##. (This should be obvious enough; I should be able to provide a proof if necessary.) However, as ##10^n## is not a power of 2, Carmichael's theorem tells us that ##\lambda\left(10^n\right)=\varphi\left(10^n\right)=4\cdot 10^{n-1}##, which doesn't divide ##10^n##.

Anyone know what's wrong here?