Find the number of positive integers x such that [(x/(m−1))]=[(x/(m+1))], for a particular integer m≥2. [ ] means G.I.F.


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Question Number 28186 by Tinkutara last updated on 25/Jan/18

$F i n d t h e n u m b e r o f p o s i t i v e i n t e g e r s$ $x s u c h t h a t [\frac{x}{m - 1}] = [\frac{x}{m + 1}], f o r a$ $p a r t i c u l a r i n t e g e r m ⩾ 2 .$ $[] m e a n s G . I . F .$
Commented by abdo imad last updated on 25/Jan/18

$i f 0 < x < m - 1 \Rightarrow 0 < \frac{x}{m - 1} < 1 a n d 0 < x < m + 1$ $\Rightarrow [\frac{x}{m - 1}] = [\frac{x}{m + 1}] = 0 l e t s u p p o s e x ⩾ m - 1 ⩾ 1$ $x = q (m - 1) + r a n d q, r i n t e g e r s 0 ⩽ r < m - 1$ $\frac{x}{m - 1} = q + \frac{r}{m - 1} \Rightarrow [\frac{x}{m - 1}] = q + [\frac{r}{m - 1}] = q + 0 = q$ $\frac{x}{m + 1} = \frac{q (m - 1) + r}{m + 1} = \frac{q (m + 1) - 2 q + r}{m + 1} = q + \frac{r - 2 q}{m + 1} a n d$ $[\frac{x}{m + 1}] = q + [\frac{r - 2 q}{m + 1}]$ $e \Leftrightarrow q = q + [\frac{r - 2 q}{m + 1}] \Leftrightarrow [\frac{r - 2 q}{m + 1}] = 0 \Leftrightarrow 0 ⩽ \frac{r - 2 q}{m + 1} < 1$ $\Leftrightarrow 0 ⩽ r - 2 q < m + 1 b u t x = q m - q + r = q (m + 1) + r - 2 q$ $\Rightarrow r - 2 q = x - q (m + 1) \Rightarrow 0 ⩽ x - q (m + 1) < m + 1$ $\Rightarrow q (m + 1) ⩽ x < (q + 1) (m + 1) . . . .$
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