MHB Can You Prove the Polynomial Has Three Distinct Roots?

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The discussion centers on a problem of proving that a polynomial \( P(x) = x^3 + ax^2 + bx + c \) with coefficients \( a, b, c \) from the set \( M = \{-10, -9, \ldots, 10\} \) has three distinct roots, given the condition \( |P(2 + \sqrt{2})| < \frac{9}{2018} \). Participants explore the implications of the polynomial's behavior at the specified point and analyze the conditions under which distinct roots can be established. The lack of responses to a previous problem of the week indicates a potential difficulty or complexity in the current problem. The conversation may reference solutions to similar problems for guidance. Ultimately, the goal is to demonstrate the distinctness of the roots based on the polynomial's characteristics.
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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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No one answered POTW #493.

However, you can refer to the solution of other below:
$P(2+\sqrt{2})=20+14\sqrt{2}+a(6+4\sqrt{2})+b(2+\sqrt{2})+c=(20+6a+2b+c)+(14+4a+b)\sqrt{2}$.

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

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