Can You Prove the Polynomial Has Three Distinct Roots?

[anemone]

Here is this week's POTW:

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Let $M={-10,\,-9,\,-8,\,\cdots,\,9,\,10}$. There exists a polynomial $P(x)=x^3+ax^2+bx+c$ with $a,\,b,\,c \in M$. Given that $|P(2+\sqrt{2}|<\dfrac{9}{2018}$. Prove that $P(x)$ has 3 distinct roots.

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[anemone]

No one answered POTW #493.

However, you can refer to the solution of other below:

Spoiler

$P(2+\sqrt{2})=20+14\sqrt{2}+a(6+4\sqrt{2})+b(2+\sqrt{2})+c=(20+6a+2b+c)+(14+4a+b)\sqrt{2}$.

Let $x=20+6a+2b+c$ and $y=14+4a+b$.

Then $|x+y\sqrt{2}|<\dfrac{9}{2018}$.

We have $-70\le x \le 110$ and $-36\le y \le 64$ and so

$|x+y\sqrt{2}| \le |x|+|y|\sqrt{2} \le 110+64\sqrt{2}<201$ and so

$|x-2y^2|\le \dfrac{9(201)}{2018}<1$.

But $x$ and $y$ are integers so $x^2-2y^2=0$. This implies $x=y=0$.

Then we have $20+6a+2b+c=0,\,14+4a+b=0$

So $b=-4a-14$ and $c=2a+8$ and that

$\begin{align*}P(x)&=x^3+ax^2-(4a+14)x+2a+8\\&=(x-2-\sqrt{2})(x-2+\sqrt{2})(x+a+4)\end{align*}$