Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 28449 by A1B1C1D1 last updated on 25/Jan/18

Answered by $@ty@m last updated on 26/Jan/18

there are infinitely many solutions: (0,±1) (±1,0) (±(3/5),±(4/5)) (±(5/(13)),±((12)/(13))) ... ... ...

$${there}\:{are}\:{infinitely}\:{many}\:{solutions}: \\ $$$$\left(\mathrm{0},\pm\mathrm{1}\right) \\ $$$$\left(\pm\mathrm{1},\mathrm{0}\right) \\ $$$$\left(\pm\frac{\mathrm{3}}{\mathrm{5}},\pm\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\left(\pm\frac{\mathrm{5}}{\mathrm{13}},\pm\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$$$... \\ $$$$... \\ $$$$... \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 26/Jan/18

General Solution: m,n∈Z, m & n are not both 0 x=((m^2 −n^2 )/(m^2 +n^2 )) , y=((2mn)/(m^2 +n^2 )) Proof: x^2 +y^2 =(((m^2 −n^2 )/(m^2 +n^2 )))^2 +(((2mn)/(m^2 +n^2 )))^2 =(((m^2 −n^2 )^2 +(2mn)^2 )/((m^2 +n^2 )^2 )) =((m^4 −2m^2 n^2 +n^4 +4m^2 n^2 )/((m^2 +n^2 )^2 )) =((m^4 +2m^2 n^2 +n^4 )/((m^2 +n^2 )^2 )) =(((m^2 +n^2 )^2 )/((m^2 +n^2 )^2 )) =1

$$\mathrm{General}\:\mathrm{Solution}: \\ $$$$\mathrm{m},\mathrm{n}\in\mathbb{Z},\:\mathrm{m}\:\&\:\mathrm{n}\:\mathrm{are}\:\mathrm{not}\:\mathrm{both}\:\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:,\:\mathrm{y}=\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{Proof}: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\left(\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{2mn}\right)^{\mathrm{2}} }{\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{m}^{\mathrm{4}} −\mathrm{2m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{4}} +\mathrm{4m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} }{\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{m}^{\mathrm{4}} +\mathrm{2m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{4}} }{\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1} \\ $$

Commented by A1B1C1D1 last updated on 26/Jan/18

Thanks.

$$\mathrm{Thanks}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com