Find all positive integral values of $x$ if $\prod_{m=0}^{1008} (x-{2m+1 \over 2})^{2m+1} \lt 0$ Lately, I have been taking multiple classes on such math problems. So while I was solving some math problems, I came over this question. The question originally says: How many positive integer solutions has the following inequality: $$\left(x-{1\over 2}\right)^1\left(x-{3\over 2}\right)^3\cdots\left(x-{2017\over 2}\right)^{2017} \lt 0.$$ I managed to change to: $$\prod_{m=0}^{1008} \left(x-{2m+1 \over 2}\right)^{2m+1} \lt 0.$$ I have been trying so hard to simplify more to get a solution but failed, and I was hoping if I could receive some help on this one. Thank you anyways.
 A: Hint : 
The inequality is equivalent to 
$$\left(x-{1\over 2}\right)\left(x-{3\over 2}\right)...\left(x-{2017\over 2}\right)\lt 0\tag1$$
(why?)
Let $f(x)$ be the LHS of $(1)$, and consider the graph of $y=f(x)$. The degree of $y=f(x)$ is $\frac{2017+1}{2}=1009$ which is odd.
So, positive integer solutions are

$2,4,\cdots, 1008$.

A: This answers the original version of the question.
Hint: $$\deg\left(\prod_{m=0}^{1008} \Big(x-{2m+1 \over 2}\Big)^{2m+1}\right) = \sum_{m=0}^{1008} (2m + 1) \equiv \sum_{m=0}^{1008} 1 \equiv 1\pmod2,$$
so the given polynomial has odd degree.

 Since a polynomial of odd degree will "explode at infinity" (i.e. $p(x)\to-\infty$ as $x\to-\infty$), but a polynomial can only have finitely many roots, it will be "eventually" negative for sufficiently large $-M$ ($M > 0$), so the answer is infinitely many.

A: $$ 
\begin{align} 
f(x)&=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m+1} \\ 
&=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m} \, \text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ 
&=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ 
&=\text{sgn}\prod_{m=0}^{x-1}\left(x-m-\frac12\right) \, \text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) \\ 
&=\text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) =\prod_{m=x}^{1008}(-1)=\color{red}{(-1)^{1009-x}\space\colon\,x\in[0,1008]} 
\end{align} 
$$
