I have the following problem, and I'm a little stuck on how I go from the given hypothesis to the conclusion. I'm not too sure how I can alter the term in the (mod) without altering everything else.
Prove that if $$a^{\phi(b)} + b^{\phi(a)}\equiv1\pmod{ab}$$ then
$$a^{\phi(b)} + b^{\phi(a)}\equiv 1+ab\pmod{2ab}$$
I ended up using the definition of mods where
$$ab\ |\ (a^{\phi(b)} + b^{\phi(a)}) - 1$$
but I couldn't figure out how to use this to get to the conclusion and I also don't understand how I can alter the stuff inside the mod without altering everything else. Thank you in advance for helping me out with this!
This statement is false. Take $a=2$ and $b=5$. Then $\phi(a) = 1$ and $\phi(b)=4$. We have $2^4 + 5^1 = 21 \equiv 1 \pmod{10}$, but $21 \not\equiv 1 + 10 \pmod{20}$.