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


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

$x^{2} - y^{2} = 2023 x, y \in N$ $H o w m a n y p a i r o f (x, y)$
	Answered by BaliramKumar last updated on 19/Mar/23
	$solution:− x^2 −y^2 =2023 d = 2023 d = 2023 = 7^1 ×17^2 total factor of 2023 = (1+1)(2+1) = 6 No. of pairs (x, y) = ⌊(6/2)⌋ = 3 {: ((2023 = 1×2023)),((2023 = 7×289)),((2023 = 17×119)) } 3 pair { (((1012, 1011))),(((148, 141))),(((68, 51))) :} x = ((2023+1)/2) = 1012, y = ((2023−1)/2) = 1011 x = ((289+7)/2) = 148, y = ((289−7)/2) = 141 x = ((119+17)/2) = 68, y = ((119−17)/2) = 51 Pair =⌊∣((No. of even factor−No. of odd factor)/2)∣⌋$
	$s o l u t i o n : -$ $x^{2} - y^{2} = 2023$ $d = 2023$ $d = 2023 = 7^{1} \times 17^{2}$ $t o t a l f a c t o r o f 2023 = (1 + 1) (2 + 1) = 6$ $N o . o f p a i r s (x, y) = ⌊ \frac{6}{2} ⌋ = 3$ $\begin{matrix} 2023 = 1 \times 2023 \\ 2023 = 7 \times 289 \\ 2023 = 17 \times 119 \end{matrix}} 3 p a i r {\begin{cases} (1012, 1011) \\ (148, 141) \\ (68, 51) \end{cases}$ $x = \frac{2023 + 1}{2} = 1012, y = \frac{2023 - 1}{2} = 1011$ $x = \frac{289 + 7}{2} = 148, y = \frac{289 - 7}{2} = 141$ $x = \frac{119 + 17}{2} = 68, y = \frac{119 - 17}{2} = 51$ $P a i r = ⌊ ∣ \frac{N o . o f e v e n f a c t o r - N o . o f o d d f a c t o r}{2} ∣ ⌋$
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