Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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 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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com