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Question Number 188753 by BaliramKumar last updated on 19/Mar/23

x^2 − y^2 = 2023 x, y ∈ N How many pair of (x, y)

$$\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}\:\:\:\:\:\:\:\:\:{x},\:{y}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{H}{ow}\:{many}\:{pair}\:{of}\:\left({x},\:{y}\right) \\ $$

Answered by BaliramKumar last updated on 19/Mar/23

solution:− x^2 −y^2 =2023 d = 2023 d = 2023 = 7^1 ×17^2 total factor of 2023 = (1+1)(2+1) = 6 No. of pairs (x, y) = ⌊(6/2)⌋ = 3 {: ((2023 = 1×2023)),((2023 = 7×289)),((2023 = 17×119)) } 3 pair { (((1012, 1011))),(((148, 141))),(((68, 51))) :} x = ((2023+1)/2) = 1012, y = ((2023−1)/2) = 1011 x = ((289+7)/2) = 148, y = ((289−7)/2) = 141 x = ((119+17)/2) = 68, y = ((119−17)/2) = 51 Pair =⌊∣((No. of even factor−No. of odd factor)/2)∣⌋

$${solution}:− \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2023}\: \\ $$$${d}\:=\:\mathrm{2023}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${d}\:=\:\mathrm{2023}\:=\:\mathrm{7}^{\mathrm{1}} ×\mathrm{17}^{\mathrm{2}} \\ $$$${total}\:{factor}\:{of}\:\mathrm{2023}\:=\:\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)\:=\:\mathrm{6} \\ $$$${No}.\:{of}\:{pairs}\:\left({x},\:{y}\right)\:=\:\lfloor\frac{\mathrm{6}}{\mathrm{2}}\rfloor\:=\:\mathrm{3}\:\:\:\:\:\:\:\: \\ $$$$\left.\begin{matrix}{\mathrm{2023}\:=\:\mathrm{1}×\mathrm{2023}}\\{\mathrm{2023}\:=\:\mathrm{7}×\mathrm{289}}\\{\mathrm{2023}\:=\:\mathrm{17}×\mathrm{119}}\end{matrix}\right\}\:\mathrm{3}\:{pair}\:\begin{cases}{\left(\mathrm{1012},\:\mathrm{1011}\right)}\\{\left(\mathrm{148},\:\mathrm{141}\right)}\\{\left(\mathrm{68},\:\mathrm{51}\right)}\end{cases} \\ $$$${x}\:=\:\frac{\mathrm{2023}+\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1012},\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{2023}−\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1011} \\ $$$${x}\:=\:\frac{\mathrm{289}+\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{148},\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{289}−\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{141} \\ $$$${x}\:=\:\frac{\mathrm{119}+\mathrm{17}}{\mathrm{2}}\:=\:\mathrm{68},\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{119}−\mathrm{17}}{\mathrm{2}}\:=\:\mathrm{51} \\ $$$$ \\ $$$${Pair}\:=\lfloor\mid\frac{{No}.\:{of}\:{even}\:{factor}−{No}.\:{of}\:{odd}\:{factor}}{\mathrm{2}}\mid\rfloor \\ $$$$ \\ $$

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