# Find the global min of $\lfloor{(1/2)(\lfloor{N/p}\rfloor+3-\sqrt{(\lfloor{N/p}\rfloor+1)^2-4N})}\rfloor$

Denote this function as $${a}_{l}$$. Here $$p$$ is prime but not necessary for the solution, just $$p \ge 2$$ is needed. This solution is for fixed $$p$$ with $$N$$ allowed to vary. Now a plot of this function shows that it oscillates until sufficiently large $$N$$. That is $$\lim_{N\rightarrow \infty} {a}_{l} = p+1$$. Also the square root term of $${a}_{l}$$ establishes that $$N\ge 2p(2p-1)$$.

The problem is to show that the first occurrence of the global minimum of $$p+1$$ occurs at $$N=p(p^2+p+1)$$. From this value of $$N$$ I can show that the global minimum is $$p+1$$. I am interested in proving this case because this value of $$N$$ is a special value of the more general set of problems that I am working on.

I have tried taking the derivative of the $${a}_{l}$$ with respect to $$k$$ with the floor functions dropped where $$N\ge p*k$$ or $$N < (k+1)p$$ from $$k \le N/p < k+1$$. The problem is that when set to zero to find the max/min the variable $$k$$ vanishes. I have also considered setting the first derivative to be less than zero and solving for $$k$$. This results in $$N=2p(2p-1)$$ which is a local max/min.

OK take $$N=p*w$$ where $$w=\lfloor{N/p}\rfloor$$. Set $${a}_{l} = p+1$$. Then we have $$\lfloor{(1/2)(w+3-2(p+1)-\sqrt{(w+1)^2-4*p*w})}\rfloor$$ Call this function inside the floor function $${a}_{l}^{\prime}$$. Then we have the condition from the floor function $$0\le {a}_{l}^{\prime} < 1$$. Solving for $$w$$ results in $$w>p^2+p$$ for $$p\ge 2$$. Thus the first valid integer solution is $$w = p^2+p+1$$ or $$N=p(p^2+p+1)$$. Substitution into the original problem indeed shows that $$p+1$$ is the global minimum with the first value at $$N=p(p^2+p+1)$$.
Once could also assume that $$N = p*w+v$$ where $$v=N \text{ mod } p$$ where $$v \in \left\{{0,1,\cdots, p-1}\right\}$$. Then you get $$w>p^2+p+v$$ where the again the first solution occurs only when $$v=0$$.