Number Theory: prove p^1-1+q^1-1 = 1 mod pq

[Jamers328]

The full problem is:

If p and q are distinct primes, prove that

p^q-1 + q^p-1 = 1 (mod pq)


I need help proving these:
p^q-1 + q^p-1 = 1 (mod p)
and
p^q-1 + q^p-1 = 1 (mod q)

then I can say that p^q-1 + q^p-1 = 1 (mod pq) is true.

I know this is a bit confusing, so please ask questions and I will clarify if you are confused. Thanks in advance.

 

[royhaas]

Well, p to any integer power is divisible by p.

 

[daon]

I need help proving these:
p^q-1 + q^p-1 = 1 (mod p)
and
p^q-1 + q^p-1 = 1 (mod q)

\(\displaystyle p^{q-1} + q^{p-1} = 1 \,\, (mod \,\, p)\) is the same as saying \(\displaystyle q^{p-1}=1 \,\, (mod \,\, p)\) since \(\displaystyle p^{q-1}\) is zero modulo p (q is prime and hence greater than 1). The latter is true by Fermat's Lil' Thm (this is an if and only if statement, so the proof reads backwards). Similarly for your second equation.

 

[Jamers328]

Well, p to any integer power is divisible by p.

That didn't help me... :?

 

[Jamers328]

I need help proving these:
p^q-1 + q^p-1 = 1 (mod p)
and
p^q-1 + q^p-1 = 1 (mod q)

\(\displaystyle p^{q-1} + q^{p-1} = 1 \,\, (mod \,\, p)\) is the same as saying \(\displaystyle q^{p-1}=1 \,\, (mod \,\, p)\) since \(\displaystyle p^{q-1}\) is zero modulo p (q is prime and hence greater than 1). The latter is true by Fermat's Lil' Thm (this is an if and only if statement, so the proof reads backwards). Similarly for your second equation.

Thank you very much.