# Prove that $\left\lfloor\frac{p^q+q^p}{pq}\right\rfloor$ is even

Let $p$ and $q$ be different primes. Prove that $$\left\lfloor\dfrac{p^q+q^p}{pq}\right\rfloor$$ is even if $p,q \neq 2$.

I wasn't sure how to use the fraction $\dfrac{p^q+q^p}{pq}$ here and how to show that the floor function is even. Also, how do we use the fact that $p$ and $q$ are prime?

Let $N = \dfrac{(p^q-p)+(q^p-q)}{pq}.$
It follows from Fermat's little theorem that $N$ is an integer, and $N$ is even because the numerator is even and the denominator is odd.
Hence $\left\lfloor\dfrac{p^q+q^p}{pq}\right\rfloor = \left\lfloor N+ \dfrac{p+q}{pq}\right\rfloor = N$, since $0 < \dfrac{p+q}{pq} < 1$.