If x^2 − y^2 = 2023^(2023) then how many pair of x,y where x, y ∈ N


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Question Number 191232 by BaliramKumar last updated on 21/Apr/23

$If x^{2} - y^{2} = 2023^{2023} then how many$ $pair of x, y w h e r e x, y \in N$
	Answered by mr W last updated on 21/Apr/23

	$2023 = 7 \times 17^{2}$ $2023^{2023} = 7^{2023} \times 17^{4046}$ $(x - y) (x + y) = a \times b = 7^{2023} \times 17^{4046}$ $x - y = a$ $x + y = b > a$ $x = \frac{a + b}{2} > y$ $y = \frac{b - a}{2}$ $a p a i r o f (a, b) \Rightarrow a p a i r o f (x, y)$ $n u m b e r o f p a i r s o f (a, b) i s$ $\frac{(2023 + 1) (4046 + 1)}{2} = 4 095 564$ $t h a t m e a n s t h e r e a r e 4 095 564 p a i r s$ $o f x, y .$
	Commented by BaliramKumar last updated on 21/Apr/23

	$Thanks Sir$
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