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Question Number 92146 by kpn last updated on 05/May/20
how do i find integers that satisfy  x^2 −y^2 =2017
$${how}\:{do}\:{i}\:{find}\:{integers}\:{that}\:{satisfy} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2017} \\ $$
Commented by Prithwish Sen 1 last updated on 05/May/20
∵ 2017 a prime  then for the soln. to be in integers we can say  x+y = 2017  x−y=1  ∴ x = 1009 amdy = 1008
$$\because\:\mathrm{2017}\:\mathrm{a}\:\mathrm{prime} \\ $$$$\mathrm{then}\:\mathrm{for}\:\mathrm{the}\:\mathrm{soln}.\:\mathrm{to}\:\mathrm{be}\:\mathrm{in}\:\mathrm{integers}\:\mathrm{we}\:\mathrm{can}\:\mathrm{say} \\ $$$$\mathrm{x}+\mathrm{y}\:=\:\mathrm{2017} \\ $$$$\mathrm{x}−\mathrm{y}=\mathrm{1} \\ $$$$\therefore\:\mathrm{x}\:=\:\mathrm{1009}\:\mathrm{amdy}\:=\:\mathrm{1008} \\ $$
Answered by mr W last updated on 05/May/20
(x−y)(x+y)=2017=prime  x−y=1  x+y=2017  ⇒x=((2017+1)/2)=1009  ⇒y=((2017−1)/2)=1008
$$\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{2017}={prime} \\ $$$${x}−{y}=\mathrm{1} \\ $$$${x}+{y}=\mathrm{2017} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2017}+\mathrm{1}}{\mathrm{2}}=\mathrm{1009} \\ $$$$\Rightarrow{y}=\frac{\mathrm{2017}−\mathrm{1}}{\mathrm{2}}=\mathrm{1008} \\ $$

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