1

Let $a$ and $b$ be relatively prime natural numbers greater than or equal to 2. Prove that $a^{\phi(b)} + b^{\phi(a)} = 1 \quad \pmod {ab}$.

The only equations I know with $\phi$ are

$a^{\phi(m)} = 1 \pmod m,$ when $a$ and $m$ are relatively prime.

$\phi(m) = (p - 1)(q - 1)$, where $m = pq$, and $p$ and $q$ are distinct prime numbers. and

$\phi(p) = p - 1$, where $p$ is a prime number.

Since a and ab are not relatively prime, I can't use the first equation. Since a and b are natural numbers, I don't think I can use the second equation. And since the number $ab$ is not prime, I can't use the third equation.

Are there any other equations I could use to solve this?

charlotte
  • 456
Anonymous
  • 77
  • I keep getting that a^$\Phi$(b) + b^$\Phi$(a) = 1 + 1 = 2. I got this from: b^$\Phi$(a) = 1 (mod a) and a^$\Phi$(b) = a (mod b) then if you multiply the first by b and the second by a you get b(b^$\Phi$(a) - 1 = 0 (mod ab) and a(a^$\Phi$(b) - 1) = 0 (mod ab) – Anonymous Feb 07 '15 at 00:54
  • See http://math.stackexchange.com/questions/1100795/how-to-prove-m-phinn-phim-equiv-1-pmodmn-where-m-and-n-are-relat – lab bhattacharjee Feb 08 '15 at 14:17

1 Answers1

1

Hint $\ $ It is $\equiv 1\,$ mod $\,a\,$ and mod $\,b,\,$ hence $\equiv 1\,$ mod their lcm $=ab\,$ by CCRT.

Remark $\ $ It is simply $\ \underbrace{(\color{#0a0}0,\color{#c00}1)}_{a^{\Large \phi(b)}} + \underbrace{(1,0)}_{b^{\Large \phi(a)}}\, =\, (1,1)\ $ in $\ \Bbb Z/a \times \Bbb Z/b\,\cong\,\Bbb Z/ab,\ $ by CRT

because $\ a^{\Large \phi(b)}\equiv \color{#0a0}0\pmod a,\ a^{\Large \phi(b)}\equiv \color{#c00}1\pmod b,\,$ i.e. it is $\,\equiv (\color{#0a0}0,\color{#c00}1) \pmod{a,b}$

Bill Dubuque
  • 272,048
  • Wouldn't it be 0 + 1 + 1 + 0 = 1 + 1 = 2 ? That's what I keep getting and I don't know what I'm doing wrong. – Anonymous Feb 07 '15 at 20:45
  • @Anonymous The tuples $,(0,1),$ and $,(1,0),$ denote elements of the product ring $,\Bbb Z/a \times \Bbb Z/b.,$ If you haven't learned about them yet then you can safely ignore the remark and use the more elementary approach in the hint. – Bill Dubuque Feb 07 '15 at 20:48