Asked by Anonymous
Prove that if gcd(m,n)=1 then m^phi(n)+n^phi(m)=congruent 1 (mod mn)
Answers
Answered by
Count Iblis
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.
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.
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