Why it is enough to prove that (m^b )^a = m(mod p)
and (m^b )^a = m(mod q) when n = p x q are prime rather than (m^b )^a = m(mod n)?

(One version of) Chinese Remainder Theorem states that, if m and n are coprime, then the congruence

x º a (mod m), x º b (mod n)

has a unique solution (mod p q).

[Another version is a generalisation to a set of pairwise coprime integers {m₁ , m₂ , ¼, m_n }, the congruence

x º a_i (mod m_i ), 1 £ i £ n

has a unique solution (mod m₁, m₂, ..., m_n )] Now, as p and q are distinct prime, they are coprime. Hence, if we let x=(m^b)^a, then we have the congruence

x º m (mod p), x º m (mod q)

Now we can immediately spot a value of x that satisfy this condition, namely x=m. Hence, x º m (mod p q), as required.

Kerwin

If (m^b )^a =m (mod p)=m (mod q). So m (mod p) =m (mod q) as p and q prime, highest common factor =1.

So, say m (mod p)= m (mod q) = r, so m= cp + r=kq + r. Solving for c and k (to some degree), cp=kq since LCF=1 c must be a multiple of q and k a multiple of p. So, m= spq +r =sn+r so m (mod n)=r=(m^b )^a .

I hope that makes sense!