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 $0 < \dfrac{p+q}{pq} < 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.