Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 189531 by cortano12 last updated on 18/Mar/23

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{pairs}\mathrm{of}\mathrm{natural}$$$$\mathrm{numbers}(\mathrm{x},\mathrm{y})\mathrm{satisfy}$$$${\mathrm{x}}^2\mathrm{y}+\gcd (\mathrm{x},{\mathrm{y}}^2)=2023$$

Answered by aleks041103 last updated on 19/Mar/23

$$letd=gcd(x,y)$$$$\Rightarrow d\mid x,yandd\mid gcd(x,y^2)$$$$\Rightarrow d\mid 2023=7\times {17}^2$$$$x=ad,y=bd,gcd(a,b)=1$$$$\Rightarrow d^3{a}^{2}b+dgcd(a,b^{2}d)=2023$$$$\Rightarrow d^3<d^3{a}^{2}b<2023\Rightarrow d<\sqrt[3]{2023}\approx 12.65$$$$\Rightarrow d⩽12andd\mid 2023=7.{17}^2$$$$\Rightarrow d=1;7$$$$$$$$1.d=1$$$$\Rightarrow gcd(x,y^2)=1$$$$\Rightarrow a^{2}b+1=2023$$$$\Rightarrow a^{2}b=2022=2.3.337\Rightarrow nosoltn.$$$$$$$$2.d=7$$$$\Rightarrow 49a^{2}b+gcd(a,7b^2)={17}^2=289$$$$gcd(a,b)=1\Rightarrow gcd(a,7b^2)=gcd(a,7)$$$$\Rightarrow 49a^{2}b+gcd(a,7)=289={17}^2$$$$gcd(a,7)=1;7$$$$ifgcd(a,7)=7\Rightarrow 7(7a^{2}b+1)={17}^2\Rightarrow nosolution$$$$\Rightarrow gcd(a,7)=1$$$$\Rightarrow 49a^{2}b=289-1=288$$$$but7\nmid 288\Rightarrow nosolution$$$$$$$$\Rightarrow thereisnosolution...$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com