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Question Number 202677 by BaliramKumar last updated on 31/Dec/23

If x^2 − y^2 = 2023^(2024) & x, y ∈ N how many set{x, y}

Ifx2−y2=20232024&x,y∈Nhowmanyset{x,y}

Answered by witcher3 last updated on 31/Dec/23

2023^(2024) =17^(4048) .7^(2024) (x+y)(x−y)=2023^(2024) { ((x+y=17^a .7^b )),((x−y=17^(4048−a) 7^(2024−b) )) :} solve in Z ⇒2x=17^a .7^b +17^(4048−a) 7^(2024−b) existe solution if and only if (a,b)≠{(0,0);(4048,2024)} we have all set of type {(x,y);(x−y)},(x,y)∈N^2 a∈[0,4048],b∈[0,2024] B=card{x,−y}=card{x,y}=A A∩B={2023^(1012) ,0} card(A∪B)=2card(A)−1 2.card{x,y}−1=(4049).(2025)−2 card{x,y}=(((4049.2025−1))/2)=4099612

20232024=174048.72024(x+y)(x−y)=20232024{x+y=17a.7bx−y=174048−a72024−bsolveinZ⇒2x=17a.7b+174048−a72024−bexistesolutionifandonlyif(a,b)≠{(0,0);(4048,2024)}wehaveallsetoftype{(x,y);(x−y)},(x,y)∈N2a∈[0,4048],b∈[0,2024]B=card{x,−y}=card{x,y}=AA∩B={20231012,0}card(A∪B)=2card(A)−12.card{x,y}−1=(4049).(2025)−2card{x,y}=(4049.2025−1)2=4099612

Answered by BaliramKumar last updated on 01/Jan/24

2023^(2024) = (7×17^2 )^(2024) = 7^(2024) ×17^(4048) No. of set{x, y}= ⌊∣((No. of even factors − No. of odd factors)/2)∣⌋ No. of set{x, y}= ⌊∣((0 − (2024+1)(4048+1))/2)∣⌋ No. of set{x, y}= ⌊∣((− (2025)(4049))/2)∣⌋ No. of set{x, y}= ⌊((8199225)/2)⌋ No. of set{x, y}= ⌊4099612.5⌋ No. of set{x, y}= 4099612

20232024=(7×172)2024=72024×174048No.ofset{x,y}=⌊∣No.ofevenfactors−No.ofoddfactors2∣⌋No.ofset{x,y}=⌊∣0−(2024+1)(4048+1)2∣⌋No.ofset{x,y}=⌊∣−(2025)(4049)2∣⌋No.ofset{x,y}=⌊81992252⌋No.ofset{x,y}=⌊4099612.5⌋No.ofset{x,y}=4099612

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