Obviously this holds for $k = 0$ and $k = 1$ sine $p - 1$ is even. The limit on $k$ is about $0.38 p$. This limit is due to the problem details however a general solution if possible for $k \le p - 2$ is just as good.
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This question is very confusing - the ratio is rational for all prime $p$ and positive $k$. Do you mean to ask whether it's integral? Are you looking for a method to determine when the result is integral ('divide them out'), or a way of counting the number of solutions (which will vary quite a bit), or what? What's the source of the upper bound on $k$? – Steven Stadnicki Apr 13 '17 at 01:01
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Yes when is this integral and when it is rational. For example when $p = 13$ then $p - 1$ is 12 and is divisible when $k = 0, 1, 2, 3, 5$ which covers the maximum values of $k = 4$. When $p = 11$ then we have $10/(k+1)$ which is divisible when $k = 0, 1$ but not when $k = 2, 3$ which covers all the cases due to the ${k}_{max}$ limit. – Lorenz H Menke Apr 13 '17 at 01:06
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The source of the upper bound on $k$ is from counting the number of factorable quadratics of the form ${p}^{2}, {x}^{2} + s, x + t = \left({p, x + a}\right) \left({p, x + b}\right)$ where $- N \le s, t \le + N$ for height $N \ge {p}^{2}$. There are correction terms to this sum which is the souce of the $k$ bound. – Lorenz H Menke Apr 13 '17 at 01:11
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Lorenz, please look at the title you typed. It says rational. – Will Jagy Apr 13 '17 at 01:17
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I assume you want the set/floor in the title to contain $p$ somehow? Right now it reads "$k \in 0$", since $\lfloor \rm that\ thing \rfloor = 0$. – pjs36 Apr 13 '17 at 03:52