Question Number 37627 by Joel579 last updated on 16/Jun/18
$$\mathrm{Given}\:\left\{{x}\right\}\:=\:{x}\:−\:\lfloor{x}\rfloor \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{from}\:\mathrm{equation} \\ $$$$\left\{{x}\right\}\:+\:\left\{{x}^{\mathrm{2}} \right\}\:=\:\mathrm{1} \\ $$$$\mathrm{with}\:−\mathrm{10}\:\leqslant\:{x}\:\leqslant\:\mathrm{10}\:? \\ $$
Answered by ajfour last updated on 16/Jun/18
![let x=m+h where m∈Z and h={x} {x}+{x^2 }=1 ⇒ h+(m+h)^2 −[(m+h)^2 ]=1 h+m^2 +2mh+h^2 −m^2 −[2mh+h^2 ]=1 ⇒ {2mh+h^2 }=1−h ⇒ h≠ 0 ...](https://www.tinkutara.com/question/Q37641.png)
$${let}\:{x}={m}+{h} \\ $$$${where}\:{m}\in\mathbb{Z}\:{and}\:{h}=\left\{{x}\right\} \\ $$$$\left\{{x}\right\}+\left\{{x}^{\mathrm{2}} \right\}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:{h}+\left({m}+{h}\right)^{\mathrm{2}} −\left[\left({m}+{h}\right)^{\mathrm{2}} \right]=\mathrm{1} \\ $$$$\:\:{h}+{m}^{\mathrm{2}} +\mathrm{2}{mh}+{h}^{\mathrm{2}} −{m}^{\mathrm{2}} −\left[\mathrm{2}{mh}+{h}^{\mathrm{2}} \right]=\mathrm{1} \\ $$$$\Rightarrow\:\left\{\mathrm{2}{mh}+{h}^{\mathrm{2}} \right\}=\mathrm{1}−{h} \\ $$$$\Rightarrow\:{h}\neq\:\mathrm{0} \\ $$$$… \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jun/18
$$\mathcal{N}{ice}\:{begining}! \\ $$
Answered by ajfour last updated on 16/Jun/18
$$\mathrm{172}\:{real}\:{solutions},\:{i}\:{think}. \\ $$
Answered by MJS last updated on 16/Jun/18
![for x>0 we must solve equations x∈[0;1[ (x−0)+(x^2 −0)=1 ⇒ x^2 +x−1=0 x∈[1;2[ (x−1)+(x^2 −1)=1 ⇒ x^2 +x−3=0 (x−1)+(x^2 −2)=1 ⇒ x^2 +x−4=0 (x−1)+(x^2 −3)=1 ⇒ x^2 +x−5=0 x∈[2;3[ (x−2)+(x^2 −4)=1 ⇒ x^2 +x−7=0 ... (x−2)+(x^2 −8)=1 ⇒ x^2 +x−11=0 x∈[n;n+1] (x−n)+(x^2 −n^2 )=1 ⇒ x^2 +x−(n^2 +n+1)=0 ... (x−n)+(x^2 −((n+1)^2 −1))=1 ⇒ ⇒ x^2 +x−(n^2 +3n+1)=0 we have 1+3+5+7+9+11+13+15+17+19= = 100 solutions in [0;10] for x<0 the negative solutions of above equations count ⇒ we have no solution in [−1;0[ 1 solution in [−2;−1[ 3 solutions in [−3;−2[ ... so we have 100+100−19=181 solutions in [−10;10]](https://www.tinkutara.com/question/Q37697.png)
$$\mathrm{for}\:{x}>\mathrm{0}\:\mathrm{we}\:\mathrm{must}\:\mathrm{solve}\:\mathrm{equations} \\ $$$${x}\in\left[\mathrm{0};\mathrm{1}\left[\right.\right. \\ $$$$\left({x}−\mathrm{0}\right)+\left({x}^{\mathrm{2}} −\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}\in\left[\mathrm{1};\mathrm{2}\left[\right.\right. \\ $$$$\left({x}−\mathrm{1}\right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)+\left({x}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)+\left({x}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{5}=\mathrm{0} \\ $$$${x}\in\left[\mathrm{2};\mathrm{3}\left[\right.\right. \\ $$$$\left({x}−\mathrm{2}\right)+\left({x}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{7}=\mathrm{0} \\ $$$$… \\ $$$$\left({x}−\mathrm{2}\right)+\left({x}^{\mathrm{2}} −\mathrm{8}\right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\mathrm{11}=\mathrm{0} \\ $$$$ \\ $$$${x}\in\left[{n};{n}+\mathrm{1}\right] \\ $$$$\left({x}−{n}\right)+\left({x}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)=\mathrm{0} \\ $$$$… \\ $$$$\left({x}−{n}\right)+\left({x}^{\mathrm{2}} −\left(\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right)\right)=\mathrm{1}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{x}^{\mathrm{2}} +{x}−\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+\mathrm{13}+\mathrm{15}+\mathrm{17}+\mathrm{19}= \\ $$$$=\:\mathrm{100}\:\mathrm{solutions}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{10}\right] \\ $$$$ \\ $$$$\mathrm{for}\:{x}<\mathrm{0}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{above} \\ $$$$\mathrm{equations}\:\mathrm{count} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{in}\:\left[−\mathrm{1};\mathrm{0}\left[\right.\right. \\ $$$$\:\:\:\:\:\:\mathrm{1}\:\mathrm{solution}\:\mathrm{in}\:\left[−\mathrm{2};−\mathrm{1}\left[\right.\right. \\ $$$$\:\:\:\:\:\:\mathrm{3}\:\mathrm{solutions}\:\mathrm{in}\:\left[−\mathrm{3};−\mathrm{2}\left[\right.\right. \\ $$$$… \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{100}+\mathrm{100}−\mathrm{19}=\mathrm{181}\:\mathrm{solutions} \\ $$$$\mathrm{in}\:\left[−\mathrm{10};\mathrm{10}\right] \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jun/18
$$\:\mathcal{N}\mathrm{ice}\:\mathrm{idea}\:\mathrm{of}\:\mathrm{interval}-\mathrm{wise}\:\mathrm{analysis}! \\ $$