The problem:
For all positive integers $a$ and $b$, if $(a, b)=1$, then $(a+b)^{\phi(a)\phi(b)}\equiv(a^{\phi(b)}+b^{\phi(a)}) \pmod{ab}$.
My work thus far.
I know that by using Euler's theorem I can show that $(a^{\phi(b)}+b^{\phi(a)})\equiv1 \pmod{ab}$.
I am currently stuck trying to show that $(a+b)^{\phi(a)\phi(b)}\equiv1\pmod{ab}$, and then putting that with my idea above to complete the proof.
Can anyone give me any hints or let me know if I am on the right track? Thanks!
Best Answer
$\!\bmod \color{#c00}a\!:\,\ c:= (\color{#c00}a\!+\!b)^{\phi(b)\phi(a)}\equiv (b^{\phi(b)})^{\phi(a)}\equiv 1\ $ by Euler-phi, by $\,\gcd(b^{\phi(b)},a)=1$
By symmetry, also $\,c\equiv 1\pmod{\!b},\,$ thus $\,c\equiv 1\pmod{\!ab}\,$ by CCRT
Remark $ $ Equivalently, by Euler, $\,\rm LHS-RHS \equiv 1-1\equiv 0\,$ both mod $\,a\,$ & $b,\,$ so it is divisible by $a$ and $b,\,$ so also by their lcm $= ab$, by $\,a,b\,$ coprime, i.e. $\,{\rm LHS\equiv RHS}\pmod{\!ab}$