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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.

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  • $\begingroup$ Since this is true for any negative integer, $x$, the answer is infinite. Perhaps you meant to ask how many positive integer solutions there are? $\endgroup$ – robjohn Sep 28 '18 at 16:48
  • $\begingroup$ @robjohn Yes that is what I meant, I forgot to include it. $\endgroup$ – user587054 Sep 29 '18 at 11:26
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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

$2,4,\cdots, 1008$.

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  • $\begingroup$ OP ask positive integers. $\endgroup$ – Takahiro Waki Sep 29 '18 at 16:07
  • $\begingroup$ @Takahiro Waki : Thanks. I've edited it. $\endgroup$ – mathlove Sep 29 '18 at 16:10
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$$ \begin{align} f(x)&=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m+1} \\ &=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m} \, \text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=0}^{x-1}\left(x-m-\frac12\right) \, \text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) =\prod_{m=x}^{1008}(-1)=\color{red}{(-1)^{1009-x}\space\colon\,x\in[0,1008]} \end{align} $$

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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.

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