Question Number 28186 by Tinkutara last updated on 25/Jan/18
![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.](https://www.tinkutara.com/question/Q28186.png)
$${Find}\:{the}\:{number}\:{of}\:{positive}\:{integers} \\ $$$${x}\:{such}\:{that}\:\left[\frac{{x}}{{m}−\mathrm{1}}\right]=\left[\frac{{x}}{{m}+\mathrm{1}}\right],\:{for}\:{a} \\ $$$${particular}\:{integer}\:{m}\geqslant\mathrm{2}. \\ $$$$\left[\:\right]\:{means}\:{G}.{I}.{F}. \\ $$
Commented by abdo imad last updated on 25/Jan/18
![if 0<x<m−1⇒0< (x/(m−1))<1 and 0<x<m+1 ⇒ [(x/(m−1))]=[(x/(m+1))]=0 let suppose x ≥m−1≥1 x=q(m−1) +r and q ,r integers 0≤r<m−1 (x/(m−1))=q +(r/(m−1)) ⇒[(x/(m−1))]=q+[(r/(m−1))]=q+0=q (x/(m+1))=((q(m−1)+r)/(m+1)) = ((q(m+1)−2q +r)/(m+1))=q + ((r−2q)/(m+1)) and [(x/(m+1))]=q +[((r−2q)/(m+1))] e ⇔q=q +[((r−2q)/(m+1))]⇔ [((r−2q)/(m+1))]=0 ⇔0≤ ((r−2q)/(m+1))<1 ⇔0≤ r−2q<m+1 but x = qm−q +r=q(m+1)+r−2q ⇒r−2q=x −q(m+1)⇒0≤x −q(m+1)<m+1 ⇒q(m+1)≤x<(q+1)(m+1)....](https://www.tinkutara.com/question/Q28376.png)
$$\:{if}\:\mathrm{0}<{x}<{m}−\mathrm{1}\Rightarrow\mathrm{0}<\:\frac{{x}}{{m}−\mathrm{1}}<\mathrm{1}\:\:\:{and}\:\:\mathrm{0}<{x}<{m}+\mathrm{1} \\ $$$$\Rightarrow\:\left[\frac{{x}}{{m}−\mathrm{1}}\right]=\left[\frac{{x}}{{m}+\mathrm{1}}\right]=\mathrm{0}\:{let}\:{suppose}\:{x}\:\geqslant{m}−\mathrm{1}\geqslant\mathrm{1} \\ $$$${x}={q}\left({m}−\mathrm{1}\right)\:+{r}\:{and}\:{q}\:,{r}\:{integers}\:\mathrm{0}\leqslant{r}<{m}−\mathrm{1} \\ $$$$\frac{{x}}{{m}−\mathrm{1}}={q}\:+\frac{{r}}{{m}−\mathrm{1}}\:\Rightarrow\left[\frac{{x}}{{m}−\mathrm{1}}\right]={q}+\left[\frac{{r}}{{m}−\mathrm{1}}\right]={q}+\mathrm{0}={q} \\ $$$$\frac{{x}}{{m}+\mathrm{1}}=\frac{{q}\left({m}−\mathrm{1}\right)+{r}}{{m}+\mathrm{1}}\:=\:\frac{{q}\left({m}+\mathrm{1}\right)−\mathrm{2}{q}\:+{r}}{{m}+\mathrm{1}}={q}\:+\:\frac{{r}−\mathrm{2}{q}}{{m}+\mathrm{1}}\:{and} \\ $$$$\left[\frac{{x}}{{m}+\mathrm{1}}\right]={q}\:+\left[\frac{{r}−\mathrm{2}{q}}{{m}+\mathrm{1}}\right]\:\: \\ $$$${e}\:\Leftrightarrow{q}={q}\:+\left[\frac{{r}−\mathrm{2}{q}}{{m}+\mathrm{1}}\right]\Leftrightarrow\:\left[\frac{{r}−\mathrm{2}{q}}{{m}+\mathrm{1}}\right]=\mathrm{0}\:\Leftrightarrow\mathrm{0}\leqslant\:\frac{{r}−\mathrm{2}{q}}{{m}+\mathrm{1}}<\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{0}\leqslant\:{r}−\mathrm{2}{q}<{m}+\mathrm{1}\:\:{but}\:{x}\:=\:{qm}−{q}\:+{r}={q}\left({m}+\mathrm{1}\right)+{r}−\mathrm{2}{q} \\ $$$$\Rightarrow{r}−\mathrm{2}{q}={x}\:−{q}\left({m}+\mathrm{1}\right)\Rightarrow\mathrm{0}\leqslant{x}\:−{q}\left({m}+\mathrm{1}\right)<{m}+\mathrm{1} \\ $$$$\Rightarrow{q}\left({m}+\mathrm{1}\right)\leqslant{x}<\left({q}+\mathrm{1}\right)\left({m}+\mathrm{1}\right)…. \\ $$$$ \\ $$$$ \\ $$