Find $x\in\mathbb Z$ to $x-4\mid x\sqrt{x}-2x$

Question

Find $x\in\mathbb Z$ to $x-4\mid x\sqrt{x}-2x$. This is my idea:

Because $x\in\mathbb Z \Rightarrow x-4\in\mathbb Z$, so $x\sqrt{x}-2x\in\mathbb Z\Rightarrow x\sqrt{x}+2x\in\mathbb Z$

So to $x\in\mathbb Z$ to\begin{gather*}x-4\mid x\sqrt{x}-2x\Rightarrow x-4\mid(x\sqrt{x}-2x)(x\sqrt{x}+2x)\\\Rightarrow x-4\mid x^3-4x^2\Rightarrow x-4\mid x^2(x-4)\end{gather*}(true for all $x\in\mathbb Z$). But this is clearly wrong. I hope your help.

@AndonisRyder You assumed that $x-4\vert x\sqrt{x}-2x$ is true for some $x$ and then you arrive at a contradiction. So unless there is an error in the manipulations that follow the assumption, the initial assumption is incorrect (so there is no value of $x$). I believe that the only solution is $x=0$. Also, when $x=0$, we can't follow the steps that you have mentioned since it would be equivalent to multiplying both sides of an equation by $0$. — Sathvik, May 29 '21 at 10:08
When $x\ne0$, the following steps are fine but we arrive at a contradiction. So our initial assumption (that there exists some $x$) is incorrect. — Sathvik, May 29 '21 at 10:14

Sathvik · Accepted Answer · 2021-05-29T09:36:11.683

2

Since $x\in \mathbb{Z}$, $$x-4\vert x\sqrt{x}-2x \implies x\sqrt{x}-2x\in \mathbb{Z}\implies x\sqrt{x}\in \mathbb{Z}\implies x=p^2,\; p\in\mathbb{Z}$$ So, we have, $$p^2-4\vert p^3-2p^2\implies p+2\vert p^2\implies p+2\vert p^2+4p+8\implies p+2\vert(p+2)^2+4\\ \therefore\; p+2\vert4$$ Therefore, $$p\in\{2,0,-1,-3,-4,-6\}\implies x\in\{4,0,1,9,16,36\} $$ Plug the above values in the original divisibility condition and check which of these work.

edited May 29 '21 at 09:36

answered May 29 '21 at 09:31

Sathvik

3,609

thank you very much – AndonisRyder May 29 '21 at 10:07
You're welcome :) – Sathvik May 29 '21 at 10:09
why i tried again only 0.4 is correct? – AndonisRyder May 29 '21 at 10:19
When $x=4$, you get $0$ divides $0$ and when $x=0$, you get $-4$ divides $0$ and these are the only solutions. – Sathvik May 29 '21 at 10:23
So why do you find so many other solutions? – AndonisRyder May 29 '21 at 10:25
They are not all solutions. The valid solutions will belong to that set and you will have to verify if they work or not. – Sathvik May 29 '21 at 10:27

Find $x\in\mathbb Z$ to $x-4\mid x\sqrt{x}-2x$

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