Question-28449

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

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

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