If-x-2-y-2-2023-2024-amp-x-y-N-how-many-set-x-y-

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}

$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2024}} \:\&\:{x},\:{y}\:\in\:\boldsymbol{\mathrm{N}} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{set}\left\{{x},\:{y}\right\} \\ $$

Answered by AST last updated on 31/Dec/23

x^2 −y^2 ≡7^(2024) ×17^(4048) =(x−y)(x+y) ⇒x−y=7^a 17^b ∧x+y=7^(2024−a) 17^(4048−b) ⇒(x,y)=(((7^(2024−a) 17^(4048−b) +7^a 17^b )/2),((7^(2024−a) 17^(4048−b) −7^a 17^b )/2)) such that x≥y ⇒a∈{0,1,2,...,2024}; b∈{0,1,2,...,4048}⇒There are ((4049×2025−1)/2)

$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \equiv\mathrm{7}^{\mathrm{2024}} ×\mathrm{17}^{\mathrm{4048}} =\left({x}−{y}\right)\left({x}+{y}\right) \\ $$$$\Rightarrow{x}−{y}=\mathrm{7}^{{a}} \mathrm{17}^{{b}} \:\wedge{x}+{y}=\mathrm{7}^{\mathrm{2024}−{a}} \mathrm{17}^{\mathrm{4048}−{b}} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\frac{\mathrm{7}^{\mathrm{2024}−{a}} \mathrm{17}^{\mathrm{4048}−{b}} +\mathrm{7}^{{a}} \mathrm{17}^{{b}} }{\mathrm{2}},\frac{\mathrm{7}^{\mathrm{2024}−{a}} \mathrm{17}^{\mathrm{4048}−{b}} −\mathrm{7}^{{a}} \mathrm{17}^{{b}} }{\mathrm{2}}\right) \\ $$$${such}\:{that}\:{x}\geqslant{y}\:\Rightarrow{a}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{2024}\right\}; \\ $$$${b}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{4048}\right\}\Rightarrow{There}\:{are}\:\frac{\mathrm{4049}×\mathrm{2025}−\mathrm{1}}{\mathrm{2}} \\ $$

Commented by BaliramKumar last updated on 31/Dec/23

set{x, y} = set{y, x} and x and y non zero If a=1012 & b=2024 then y=0

$$\mathrm{set}\left\{{x},\:{y}\right\}\:=\:{set}\left\{{y},\:{x}\right\}\:\:{and}\:\:\:\:{x}\:{and}\:{y}\:{non}\:{zero} \\ $$$$\mathrm{If}\:\mathrm{a}=\mathrm{1012}\:\:\:\&\:\mathrm{b}=\mathrm{2024}\:\mathrm{then}\:\mathrm{y}=\mathrm{0} \\ $$

Commented by AST last updated on 31/Dec/23

$${What}'{s}\:{the}\:{answer}? \\ $$

Commented by BaliramKumar last updated on 31/Dec/23

⌊(((2024+1)(4048+1))/2)⌋ = 4099612 Any computer programmer please check the answer

$$\lfloor\frac{\left(\mathrm{2024}+\mathrm{1}\right)\left(\mathrm{4048}+\mathrm{1}\right)}{\mathrm{2}}\rfloor\:=\:\mathrm{4099612} \\ $$$$\mathrm{Any}\:\mathrm{computer}\:\mathrm{programmer}\:\mathrm{please} \\ $$$$\mathrm{check}\:\mathrm{the}\:\mathrm{answer} \\ $$

Commented by MathematicalUser2357 last updated on 06/Jan/24

$$\mathrm{Right} \\ $$

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

$$\mathrm{2023}^{\mathrm{2024}} =\mathrm{17}^{\mathrm{4048}} .\mathrm{7}^{\mathrm{2024}} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{2023}^{\mathrm{2024}} \\ $$$$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{17}^{\mathrm{a}} .\mathrm{7}^{\mathrm{b}} }\\{\mathrm{x}−\mathrm{y}=\mathrm{17}^{\mathrm{4048}−\mathrm{a}} \mathrm{7}^{\mathrm{2024}−\mathrm{b}} }\end{cases} \\ $$$$\mathrm{solve}\:\mathrm{in}\:\mathbb{Z} \\ $$$$\Rightarrow\mathrm{2x}=\mathrm{17}^{\mathrm{a}} .\mathrm{7}^{\mathrm{b}} +\mathrm{17}^{\mathrm{4048}−\mathrm{a}} \mathrm{7}^{\mathrm{2024}−\mathrm{b}} \\ $$$$\mathrm{existe}\:\mathrm{solution}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)\neq\left\{\left(\mathrm{0},\mathrm{0}\right);\left(\mathrm{4048},\mathrm{2024}\right)\right\} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{all}\:\mathrm{set}\:\mathrm{of}\:\mathrm{type}\:\left\{\left(\mathrm{x},\mathrm{y}\right);\left(\mathrm{x}−\mathrm{y}\right)\right\},\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{N}^{\mathrm{2}} \\ $$$$\mathrm{a}\in\left[\mathrm{0},\mathrm{4048}\right],\mathrm{b}\in\left[\mathrm{0},\mathrm{2024}\right] \\ $$$$\mathrm{B}=\mathrm{card}\left\{\mathrm{x},−\mathrm{y}\right\}=\mathrm{card}\left\{\mathrm{x},\mathrm{y}\right\}=\mathrm{A} \\ $$$$\mathrm{A}\cap\mathrm{B}=\left\{\mathrm{2023}^{\mathrm{1012}} ,\mathrm{0}\right\} \\ $$$$\mathrm{card}\left(\mathrm{A}\cup\mathrm{B}\right)=\mathrm{2card}\left(\mathrm{A}\right)−\mathrm{1} \\ $$$$\mathrm{2}.\mathrm{card}\left\{\mathrm{x},\mathrm{y}\right\}−\mathrm{1}=\left(\mathrm{4049}\right).\left(\mathrm{2025}\right)−\mathrm{2} \\ $$$$\mathrm{card}\left\{\mathrm{x},\mathrm{y}\right\}=\frac{\left(\mathrm{4049}.\mathrm{2025}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{4099612} \\ $$$$ \\ $$

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

$$\mathrm{2023}^{\mathrm{2024}} \:=\:\left(\mathrm{7}×\mathrm{17}^{\mathrm{2}} \right)^{\mathrm{2024}} \:=\:\mathrm{7}^{\mathrm{2024}} ×\mathrm{17}^{\mathrm{4048}} \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\lfloor\mid\frac{\mathrm{No}.\:\mathrm{of}\:\mathrm{even}\:\mathrm{factors}\:−\:\mathrm{No}.\:\mathrm{of}\:\mathrm{odd}\:\mathrm{factors}}{\mathrm{2}}\mid\rfloor \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\lfloor\mid\frac{\mathrm{0}\:−\:\left(\mathrm{2024}+\mathrm{1}\right)\left(\mathrm{4048}+\mathrm{1}\right)}{\mathrm{2}}\mid\rfloor \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\lfloor\mid\frac{−\:\left(\mathrm{2025}\right)\left(\mathrm{4049}\right)}{\mathrm{2}}\mid\rfloor \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\lfloor\frac{\mathrm{8199225}}{\mathrm{2}}\rfloor\: \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\lfloor\mathrm{4099612}.\mathrm{5}\rfloor \\ $$$$\mathrm{No}.\:\mathrm{of}\:\mathrm{set}\left\{\mathrm{x},\:\mathrm{y}\right\}=\:\mathrm{4099612} \\ $$$$ \\ $$

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