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

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

$$thereareinfinitelymanysolutions:$$$$(0,\pm 1)$$$$(\pm 1,0)$$$$(\pm \frac{3}{5},\pm \frac{4}{5})$$$$(\pm \frac{5}{13},\pm \frac{12}{13})$$$$...$$$$...$$$$...$$$$$$

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

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

Commented by A1B1C1D1 last updated on 26/Jan/18

$$\mathrm{Thanks}.$$

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