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Question Number 115487 by 171 last updated on 26/Sep/20

Commented by Dwaipayan Shikari last updated on 26/Sep/20

$${\mathrm{y}}^2=1+1(\mathrm{x}=1)$$$$\mathrm{y}=\pm \sqrt{2}$$$${\mathrm{y}}^2=1+2\Rightarrow \mathrm{y}=\pm \sqrt{3}$$$${\mathrm{y}}^2=1+6\Rightarrow \mathrm{y}=\pm \sqrt{7}$$$$0⩽\mathrm{x}<\mathrm{\infty }$$$$\mathrm{y}\in (-\mathrm{\infty },-\sqrt{2}]\cup [\sqrt{2},\mathrm{\infty })$$

Commented by mr W last updated on 26/Sep/20

$$ifoundfollowingintegersolutions:$$$$5^2=1+4!=25$$$${11}^2=1+5!=121$$$${71}^2=1+7!=5041$$$$i{don}^{\prime }tknowiftherearemore.$$

Commented by Olaf last updated on 26/Sep/20

$$\mathrm{With}\mathrm{the}\mathrm{plot}\mathrm{these}\mathrm{are}\mathrm{the}\mathrm{only}$$$$\mathrm{solutions}.\mathrm{But}\mathrm{for}y\mathrm{we}\mathrm{take}\mathrm{the}$$$$\mathrm{sign}\mathrm{into}\mathrm{account}.$$$$x=4,y=\pm 5$$$$x=5,y=\pm 11$$$$x=7,y=\pm 71$$

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