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Question Number 147103 by mnjuly1970 last updated on 18/Jul/21

	Answered by mr W last updated on 18/Jul/21

	$x = n + f, n \in Z, 0 ⩽ f < 1$ $x^{2} = {(n + f)}^{2} = n^{2} + f^{2} + 2 n f$ $c a s e 1 : f^{2} + 2 n f < 1$ $n (n^{2} - 1) = n^{2} + 2$ $n^{3} - n^{2} - n - 2 = 0$ $\Rightarrow n = 2$ $f^{2} + 4 f - 1 < 0$ $0 ⩽ f < - 2 + \sqrt{5}$ $\Rightarrow 2 ⩽ x < \sqrt{5}$ $c a s e 2 : 1 ⩽ f^{2} + 2 n f < 2$ $n (n^{2} + 1 - 1) = n^{2} + 1 + 2$ $n^{3} - n^{2} - 3 = 0$ $\Rightarrow n o s o l u t i o n$ $c a s e 3 : 3 ⩽ f^{2} + 2 n f < 3$ $n (n^{2} + 2 - 1) = n^{2} + 2 + 2$ $n^{3} - n^{2} - n - 5 = 0$ $\Rightarrow n o s o l u t i o n$ $\Rightarrow o n l y s o l u t i o n i s 2 ⩽ x < \sqrt{5}$
	Commented by mnjuly1970 last updated on 18/Jul/21

	$t h x s i r W . . g r a t e f u l . . .$
	Answered by Kamel last updated on 18/Jul/21
	$[x]([x^2 ]−1)=[x^2 ]−1+3⇔([x]−1)([x^2 ]−1)=3 3∈P⇔(1): { (([x]−1=±3)),(([x^2 ]−1=±1)) :}∨(2): { (([x]−1=±1)),(([x^2 ]−1=±3)) :} (1)⇔[x]=4∧[x^2 ]=2⇔4≤x<5 ∧ (√2)≤x<(√3) no solutions. (2)⇔[x]=2∧[x^2 ]=4⇔2≤x<(√5) ∧ 2≤x<3 ⇒2≤x<(√5). ∴ S={x∈R/2≤x<(√5)} No solutions for −1 and −3$
	$[x] ([x^{2}] - 1) = [x^{2}] - 1 + 3 \Leftrightarrow ([x] - 1) ([x^{2}] - 1) = 3$ $3 \in P \Leftrightarrow (1) : {\begin{cases} [x] - 1 = \pm 3 \\ [x^{2}] - 1 = \pm 1 \end{cases} \lor (2) : {\begin{cases} [x] - 1 = \pm 1 \\ [x^{2}] - 1 = \pm 3 \end{cases}$ $(1) \Leftrightarrow [x] = 4 \land [x^{2}] = 2 \Leftrightarrow 4 ⩽ x < 5 \land \sqrt{2} ⩽ x < \sqrt{3} n o s o l u t i o n s .$ $(2) \Leftrightarrow [x] = 2 \land [x^{2}] = 4 \Leftrightarrow 2 ⩽ x < \sqrt{5} \land 2 ⩽ x < 3 \Rightarrow 2 ⩽ x < \sqrt{5} .$ $∴ S = {x \in R / 2 ⩽ x < \sqrt{5}}$ $N o s o l u t i o n s f o r - 1 a n d - 3$
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