find element of set S = { ((x^3 −3x^2 +2)/(2x+1)) ∈ Z for x∈Z }


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Question Number 119487 by liberty last updated on 25/Oct/20
$find element of set S = { ((x^3 −3x^2 +2)/(2x+1)) ∈ Z for x∈Z }$
$f i n d e l e m e n t o f s e t S = {\frac{x^{3} - 3 x^{2} + 2}{2 x + 1} \in Z f o r x \in Z}$
	Answered by floor(10²Eta[1]) last updated on 25/Oct/20
	$2x+1∣x^3 −3x^2 +2⇔ 2x+1∣2(x^3 −3x^2 +2)−x^2 (2x+1)=−7x^2 +4 2x+1∣2(−7x^2 +4)+7x(2x+1)=7x+8 2x+1∣2(7x+8)−7(2x+1)=9 2x+1∈{±1, ±3, ±9} 2x+1=1⇒x=0 2x+1=−1⇒x=−1 2x+1=3⇒x=1 2x+1=−3⇒x=−2 2x+1=9⇒x=4 2x+1=−9⇒x=−5 S={−5, −2, −1, 0, 1, 4}$
	$2 x + 1 ∣ x^{3} - {3 x}^{2} + 2 \Leftrightarrow$ $2 x + 1 ∣ 2 (x^{3} - {3 x}^{2} + 2) - x^{2} (2 x + 1) = - {7 x}^{2} + 4$ $2 x + 1 ∣ 2 (- {7 x}^{2} + 4) + 7 x (2 x + 1) = 7 x + 8$ $2 x + 1 ∣ 2 (7 x + 8) - 7 (2 x + 1) = 9$ $2 x + 1 \in {\pm 1, \pm 3, \pm 9}$ $2 x + 1 = 1 \Rightarrow x = 0$ $2 x + 1 = - 1 \Rightarrow x = - 1$ $2 x + 1 = 3 \Rightarrow x = 1$ $2 x + 1 = - 3 \Rightarrow x = - 2$ $2 x + 1 = 9 \Rightarrow x = 4$ $2 x + 1 = - 9 \Rightarrow x = - 5$ $S = {- 5, - 2, - 1, 0, 1, 4}$
	Answered by 1549442205PVT last updated on 25/Oct/20
	$((x^3 −3x^2 +2)/(2x+1))=k⇒((8x^3 −24x^2 +16)/(2x+1))=8k(k∈Z) ⇔4x^2 −14x+7−(9/(2x+1))=8k∈Z.Since x∈Z ⇒2x+1 ∣9⇔2x+1∈{±1,±3,±9} ⇔x∈{0,−1,1,−2,4,−5}.Checking we obtain k={2,0,6,22}.Thus,S has four elements are S={0,2,6,22}$
	$\frac{x^{3} - 3 x^{2} + 2}{2 x + 1} = k \Rightarrow \frac{{8 x}^{3} - {24 x}^{2} + 16}{2 x + 1} = 8 k (k \in Z)$ $\Leftrightarrow {4 x}^{2} - 14 x + 7 - \frac{9}{2 x + 1} = 8 k \in Z . Since x \in Z$ $\Rightarrow 2 x + 1 ∣ 9 \Leftrightarrow 2 x + 1 \in {\pm 1, \pm 3, \pm 9}$ $\Leftrightarrow x \in {0, - 1, 1, - 2, 4, - 5} . Checking we$ $obtain k = {2, 0, 6, 22} . Thus, S has four$ $elements are S = {0, 2, 6, 22}$
	Commented by floor(10²Eta[1]) last updated on 25/Oct/20
	$if x=2 then ((x^3 −3x^2 +2)/(2x+1))=((−2)/5)∉Z and x=6⇒((x^3 −3x^2 +2)/(2x+1))=((−110)/(13))∉Z the solutions are {−5, −2, −1, 0, 1, 4}$
	$if x = 2 then \frac{x^{3} - {3 x}^{2} + 2}{2 x + 1} = \frac{- 2}{5} \notin Z$ $and x = 6 \Rightarrow \frac{x^{3} - {3 x}^{2} + 2}{2 x + 1} = \frac{- 110}{13} \notin Z$ $the solutions are {- 5, - 2, - 1, 0, 1, 4}$
	Commented by 1549442205PVT last updated on 25/Oct/20

	$Elements of S are the values of the$ $expression \frac{x^{3} - {3 x}^{2} + 2}{2 x + 1}$
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