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Question Number 92146 by kpn last updated on 05/May/20

$$howdoifindintegersthatsatisfy$$$$x^2-y^2=2017$$

Commented by Prithwish Sen 1 last updated on 05/May/20

$$\because 2017\mathrm{a}\mathrm{prime}$$$$\mathrm{then}\mathrm{for}\mathrm{the}\mathrm{soln}.\mathrm{to}\mathrm{be}\mathrm{in}\mathrm{integers}\mathrm{we}\mathrm{can}\mathrm{say}$$$$\mathrm{x}+\mathrm{y}=2017$$$$\mathrm{x}-\mathrm{y}=1$$$$\therefore \mathrm{x}=1009\mathrm{amdy}=1008$$

Answered by mr W last updated on 05/May/20

$$(x-y)(x+y)=2017=prime$$$$x-y=1$$$$x+y=2017$$$$\Rightarrow x=\frac{2017+1}{2}=1009$$$$\Rightarrow y=\frac{2017-1}{2}=1008$$

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