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Find the elements of the following set: $$S=\left\{x\in\mathbb{Z}\biggm\vert\frac{x^{3}-3x+2}{2x+1}\in\mathbb{Z}\right\}$$

Angelo
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Joel ivan
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  • Well, can you find a few of them? $0$ is an element as $\frac{0^3 - 3\cdot 0 + 2}{2\cdot 0 + 1} = \frac 21 = 2\in \mathbb Z$. $1$ is too because $\frac {1^3 -3\cdot 1 + 2}{2\cdot+1} =\frac 03 = 0 \in \mathbb Z$. Can you find any more. Can you generalize. Try something and tell us what you tried and we will give you more advise. But just try something. – fleablood May 07 '21 at 02:46
  • I was doing it like this but the teacher told me that it is with the division algorithm and that I had to divide 8/1 by the trinomial and find the quotient and the remainder. – Joel ivan May 07 '21 at 03:10
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    @Joel Then have you tried doing what your teacher suggested? If not, then which step are you having trouble with? – Ben Grossmann May 07 '21 at 03:12
  • Los estoy haciendo así pero porque uso 8/1 y los otros valores no se usan y cómo al final obtengo los elementos – Joel ivan May 07 '21 at 03:33
  • Have you learned about the division algorithm in class? If so, this is a straightforward application of it. – TheBestMagician May 07 '21 at 03:44
  • It is very simple if you use this method explained in the linked dupe. – Bill Dubuque May 07 '21 at 08:41

1 Answers1

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Find the elements of the following set: $$S=\left\{x\in\mathbb{Z}\biggm\vert\frac{x^3-3x+2}{2x+1}\in\mathbb{Z}\right\}\;.$$
By using the division algorithm we get that

$\begin{align} x^3-3x+2&=\left(\dfrac12x^2-\dfrac14x-\dfrac{11}8\right)\big(2x+1\big)+\dfrac{27}8=\\ &=\dfrac18\bigg[\big(4x^2-2x-11\big)\big(2x+1\big)+27\bigg]\;. \end{align}$

Hence,

$\dfrac{x^3-3x+2}{2x+1}=\dfrac18\left(4x^2-2x-11+\dfrac{27}{2x+1}\right)\;.$

By letting $\;\lambda=\dfrac{x^3-3x+2}{2x+1}\in\mathbb{Z}\;,\;$ we get that

$8\lambda=4x^2-2x-11+\dfrac{27}{2x+1}\;\;,$

$\dfrac{27}{2x+1}=8\lambda-4x^2+2x+11\in\mathbb{Z}\;,$

consequently,

$2x+1\;$ divides $\;27\;,\;$ therefore ,

$2x+1=\pm1\;\text{ or }\;2x+1=\pm3\;\text{ or }\;2x+1=\pm9\\\qquad\qquad\quad\text{ or }\;2x+1=\pm27\;.$

By solving the previous equations, we get that

$x\in\big\{0, -1, 1, -2, 4, -5, 13, -14\big\}\;.$

Moreover,

$x=0\implies\dfrac{x^3-3x+2}{2x+1}=2\in\mathbb{Z}\;\;,$

$x=-1\implies\dfrac{x^3-3x+2}{2x+1}=-4\in\mathbb{Z}\;\;,$

$x=1\implies\dfrac{x^3-3x+2}{2x+1}=0\in\mathbb{Z}\;\;,$

$x=-2\implies\dfrac{x^3-3x+2}{2x+1}=0\in\mathbb{Z}\;\;,$

$x=4\implies\dfrac{x^3-3x+2}{2x+1}=6\in\mathbb{Z}\;\;,$

$x=-5\implies\dfrac{x^3-3x+2}{2x+1}=12\in\mathbb{Z}\;\;,$

$x=13\implies\dfrac{x^3-3x+2}{2x+1}=80\in\mathbb{Z}\;\;,$

$x=-14\implies\dfrac{x^3-3x+2}{2x+1}=100\in\mathbb{Z}\;\;,$

hence,

$S=\big\{0, -1, 1, -2, 4, -5, 13, -14\big\}\;.$

Angelo
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