- If $a$, $b$ are coprime, then $$a^{\phi(b)}+b^{\phi(a)}\equiv 1 \bmod (ab) \, .$$
- If $\left(n=2\phi(n)\right)$, then $n$ is a power of $2$.
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1) It is easy to use Euler's theorem to verify that $a^{\phi(b)}+b^{\phi(a)}=1$ (mod a), similarly $a^{\phi(b)}+b^{\phi(a)}=1$ (mod b). Since a,b are coprime we get: $a^{\phi(b)}+b^{\phi(a)}=1$ (mod ab) 2)Let $n=i2^j$ (where i is odd). $\phi(n)=\phi(2^j)\phi(i)=2^{j-1}\phi(i)=n/2$, thus $n=2^j\phi(i)$. Therefore $\phi(i)=i$, hence $i=1$ |
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Hint: For (2): Since $$\varphi(n)=n\prod_{p|n}\left(1-\frac1p\right)$$ Saying that $n=2\varphi(n)$ is equivalent to $\prod_{p|n}\left(1-\frac1p\right)=\frac12$. Can you show that this implies that if $p|n$ then $p=2$? |
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