I need to prove a lemma. I have reduced the proof to the case of the following simple congruence problem:
Let $p$ and $q$ be two positive integers.
Let $a = q + \frac{p+1}{2} + \frac{(q)(q+1)}{p} = \frac{2pq + (p)(p+1)+2(q)(q+1)}{2p}$ and
$b = q + \frac{p+1}{2} - \frac{(q)(q+1)}{p} = \frac{2pq + (p)(p+1)-2(q)(q+1)}{2p}$
I claim that $a$ and $b$ are either both (proper)rational or both integer.
Equivalently, $c \cong \,0\, mod\, 2p \,\text{iff}\, d\, \cong \,0 \,mod \,2p$ where $c = (2p)a$ and $d = (2p)b$.
Let $m = 2pq + (p)(p+1)$, then $c = m + (2(q)(q+1))$ and $d = m - (2(q)(q+1))$.
Case 1 : $p$ is odd
In this case $m$ is divisible by $2p$ and the claim follows.
Case 2 : $p$ is even
I dont know, how to proceed in the case when $p$ is even.
Any hint or suggestion would be very helpful
Have a good day
Clearly from the definition, $a$ and $b$ are both rational and equidistant from a central value, $q+\frac {p+1}{2}$, which is either integer or an integer plus one-half.
This inevitably means that whatever the value of the other term, $\frac{q(q+1)}{p}$, it must take either both $a$ and $b$ to integer value, or neither. This can be shown by considering the cases described for the central value, but is most simply demonstrated by Kelenner's observation that $a+b = 2q+p+1$ is integer.