6
$\begingroup$

Assume $p$ is a prime number such that $p\equiv 1 \pmod3$, and $q=\lfloor \frac{2p}{3}\rfloor$.

If: $$\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} =\frac{m}{n}$$

For some integers $m,n$, what is the proof that $p\mid m$

  • 0
    I changed the dots to multiplications, I hope that is what you ment.2012-08-25
  • 0
    Also is it clear that $q-1$ is odd, or could the sum possibly end with $\frac{1}{q(q+1)}$?2012-08-25
  • 2
    @SimonMarkett $q-1$ is odd since $p$ is of the form $3k+1$, so $q = \lfloor (6k+2)/3 \rfloor = \lfloor 2k+2/3 \rfloor= 2k.$2012-08-25

1 Answers 1