# Phi function application

Can anyone show some hints to this? If $\gcd(a,b)=1$, then $a^{\phi(b)}+b^{\phi(a)}=1 \pmod{ab}$. I know that $a^{\phi(b)}=1 \pmod{b}$, and similarly, $b^{\phi(a)}=1 \pmod{a}$, but then how do I combine the work so that I get the result I need? Thanks!

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You have posted 8 questions so far, accepted none of them, and been posting for a week. Could you please write your posts as questions rather than orders? And if you find that your previous questions have been satisfactorily answered, then please consider accepting the answer you find most helpful. See the FAQ on accepting answers. –  Arturo Magidin Feb 17 '11 at 20:31
Thanks for advice! –  kira Feb 17 '11 at 20:32
Thanks for going through your old posts, and thank you for editing the question. Instead of looking at each summand separately, consider the entire sum $a^{\phi(b)}+b^{\phi(a)}$ modulo $a$ and modulo $b$, as Bill and I both suggested below. –  Arturo Magidin Feb 18 '11 at 4:54
@Arturo Magidin: Thank you a lot for guiding me within this wonderful math forum. –  kira Feb 18 '11 at 5:06

It's just $\rm\ (0,1) + (1,0)\: =\: (1,1)\$ in $\rm\ \mathbb Z/a \times \mathbb Z/b\:,\$ i.e. $\rm\ a^{\phi(b)}\to (0,1)\$ being $\ 0\:$ mod $\rm\:a\:,\ 1\:$ mod $\rm\: b$

This is the special case $\rm\ \alpha = \beta = 1\$ in the following form of the Chinese Remainder Theorem:

THEOREM (CRT) $\rm\ \$ If $\rm\ a,\:b\:$ are coprime integers then

$\rm\displaystyle\quad\quad\quad\quad\quad x\equiv \alpha\ (mod\ a),\ \ x\equiv \beta\ (mod\ b)\ \ \iff\ \ x\ \equiv\ \alpha\ b^{\phi(a)}+ \beta\ a^{\phi(b)} \ \ (mod\ a\:b)$

Proof $\rm\ (\Leftarrow)\ \$ Immediate from $\rm\ b^{\phi(a)}\equiv 1\ (mod\ a),\ \ a^{\phi(b)}\equiv 1\ (mod\ b)\:$.

$\rm\ (\Rightarrow)\ \$ The solution is unique $\rm\ (mod\ ab)\$ since if $\rm\ x',\:x\$ are solutions then $\rm\ x'\equiv x\$ mod $\rm\:a,b\:$ therefore $\rm\ a,\:b\ |\ x'-x\ \Rightarrow\ a\:b\ |\ x'-x\ \$ since $\rm\ \:a,\:b\:$ coprime $\rm\:\Rightarrow\ lcm(a,b) = a\:b\:.\quad\quad$ QED

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Thanks a lot! –  kira Feb 18 '11 at 5:10

Verify the congruence holds modulo $a$ and modulo $b$. Then use the fact that if $a|k$, $b|k$, and $\gcd(a,b)=1$, then $ab|k$.

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