# This appears to be in direct violation of Carmichael's theorem.

1. Mar 20, 2014

### Whovian

In an attempt to prove a statement about the residues of a certain sequence mod $10^n$, I've derived something which seems to be in direct violation of Carmichael's theorem. Of course, this can't be right, so can someone either explain what bit of my reasoning is wrong or why this isn't in violation of Carmichael's Theorem? First of all, let $\lambda$ be the Carmichael function, and let $k$ be coprime to 2 and 5.

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?

2. Mar 20, 2014

### gopher_p

According to Wikipedia, for $n\geq4$ $$\lambda(10^n)=\text{lcm}\left(\lambda(2^n), \lambda(5^n)\right)=\text{lcm}\left(\frac{1}{2}\varphi(2^n), \varphi(5^n)\right)=\ldots=5\cdot10^{n-2},$$ and everything is right with the universe?

3. Mar 24, 2014

### Whovian

Ah. "A power of an odd prime, twice the power of an odd prime, and for 2 and 4."

*Collides hand with forehead to indicate frustration with self*