show that a^phi(b) + b^phi(a) is congruent to 1 (mod ab), if a and b are relatively prime positive integers
Euler's theorem: $\displaystyle (a,b) = 1 \ \Rightarrow \ a^{\phi (b)} \equiv 1 \ (\text{mod } b)$
Clearly: $\displaystyle b \mid b^{\phi (a)} \ \Leftrightarrow \ b^{\phi (a)} \equiv 0 \ (\text{mod } b)$
So: $\displaystyle a^{\phi (b)} + b^{\phi (a)} \equiv 1 + 0 \equiv 1 \ (\text{mod } b)$
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