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Let $n$ be a positive integer. Let $\phi(n)$ be the Euler phi function and $\lambda(n)$ be the Carmichael lambda. I know that $\phi(n)=\phi(2^{t})\phi(p_1^{t_1})\cdots\phi(p_n^{t_n})$. I also know that for $\lambda(n)=LCM[\lambda(2^t),\phi(p_1^{t_1}),\cdots,\phi(p_n^{t_n})]$ and $\lambda(2^t)$ has three cases. If $t=1,2$, the $\lambda(2^t)=1,2$. If $t\geq 3$, then $\lambda(2^t)=\phi(2^t)/2$.

Numerically I've demonstrated to myself that this is true. I'm not sure how to prove it.

emka
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  • Isn't it just because $lcm(a,b)$ divides $ab$ ? – lhf Apr 11 '14 at 03:03
  • My favorite proof of this result relies on the Chinese Remainder Theorem to write $U(n)$, the group of units mod $n$, as a product of the group of units corresponding to the prime factorization of $n$. Then $\lambda(n)$ is the exponent of $U(n)$ and $\phi(n)$ is one exponent of $U(n)$ by Lagrange. Then of course $\lambda(n)$ divides $\phi(n)$. – lhf Apr 11 '14 at 03:07
  • In the course I'm in we don't have the machinery of Groups. That would have made the proof nicer. – emka Apr 11 '14 at 03:14

1 Answers1

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Argue with orders in the multiplicative group of units.

Pedro
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