imc

We perform casework on $P(n)\le 0:$ 1. $P(n)=0$ In this case, there are $100$ such integers $n:$ $$1^2,2^2,3^2,\dots ,100^2.$$ 3. $P(n)<0$ There are $100$ factors in $P(x),$ and we need an odd number of them to be negative. We construct the table below: \begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Interval of }\boldsymbol{x} & \boldsymbol{\#}\textbf{ of Negative Factors} & \textbf{Valid?} \\ [0.5ex] \hline & & \\ [-2ex] \left(-\infty,1^2\right) & 100 & \\ [0.5ex] \left(1^2,2^2\right) & 99 & \checkmark \\ [0.5ex] \left(2^2,3^2\right) & 98 & \\ [0.5ex] \left(3^2,4^2\right) & 97 & \checkmark \\ [0.5ex] \left(4^2,5^2\right) & 96 & \\ [0.5ex] \left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] \left(6^2,7^2\right) & 94 & \\ \vdots & \vdots & \vdots \\ [0.75ex] \left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] \left(100^2,\infty\right) & 0 & \\ [0.5ex] \end{array} Note that there are $50$ valid intervals of $x.$ We count the integers in these intervals: $$\begin{array}{rl}\left(2^2-1^2-1\right)+\left(4^2-3^2-1\right)+\left(6^2-5^2-1\right)+\cdots +\left(100^2-99^2-1\right)& =\underset{(2+1)(2-1)}{\underbrace{\left(2^2-1^2\right)}}+\underset{(4+3)(4-3)}{\underbrace{\left(4^2-3^2\right)}}+\underset{(6+5)(6-5)}{\underbrace{\left(6^2-5^2\right)}}+\cdots +\underset{(100+99)(100-99)}{\underbrace{\left(100^2-99^2\right)}}-50\\ & =\underset{1+2+3+4+5+6+\cdots +99+100}{\underbrace{(2+1)+(4+3)+(6+5)+\cdots +(100+99)}}-50\\ & =\frac{101(100)}{2}-50\\ & =5000.\end{array}$$ In this case, there are $5000$ such integers $n.$ Together, the answer is $100+5000=\boxed{5100}.$