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  1. If $a$, $b$ are coprime, then $$a^{\phi(b)}+b^{\phi(a)}\equiv 1 \bmod (ab) \, .$$
  2. If $\left(n=2\phi(n)\right)$, then $n$ is a power of $2$.
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(1) is a generalization of…. – lab bhattacharjee Dec 1 '12 at 14:15
up vote 2 down vote accepted

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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why $\phi(i)$=i?andi=1? – Sali Me Dec 1 '12 at 15:12
if $\phi(i)=i$, it follows that all numbers in the set {1,2,...,i} are coprime to $i$, thus $i,i$ are coprime. Hence, $i=1$ – Amr Dec 1 '12 at 15:13
Yes, which is $2^{a-1}$ – Amr Dec 1 '12 at 15:16
@Amr: Oh sorry! – Babak S. Dec 1 '12 at 15:16
if n isn't equal $i2^j$ then ...? can you solve this problem generality? – Sali Me Dec 1 '12 at 15:18

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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I cann't show that n is power of 2,please help me. – Sali Me Dec 1 '12 at 14:46
Assume that there exists $p|n$, $p\neq 2$. Then $\frac{p-1}{p}$ appears in the product. Since $p+1$ is not a prime, $p$ does not appear in the numerator in any factor. Hence $p$ remains in the denumerator in the result as well. – Dennis Gulko Dec 1 '12 at 14:51

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