Prove that if gcd(m,n)=1 then m^phi(n)+n^phi(m)=congruent 1 (mod mn)

Answer 1

If GCD(a,r) = 1 then

a^phi(r) = 1 mod r,

therefore:

m^phi(n)+n^phi(m) mod n = 1

and

m^phi(n)+n^phi(m) mod m = 1

So, mod(nm) it is 1.