A Nice Math Olympiad Algebra Problem | How to solve for X and Y in this Problem ?

[@alibarka6490]

❤❤❤❤❤❤

[@paulortega5317]

Same idea. Solve first for xy Let f(n) = x^n + y^n f(0) = 2 f(n+2)=f(1)*f(n+1) - xy*f(n) Let f(1) = a Let xy = b f(n+2)=a*f(n+1) - b*f(n) You can quickly determine f(5) = a^5 - 5*a^3*b + 5*a*b^2 Given f(5) = 464 and given f(1)=a=4, you get 20*b^2 - 320*b + 1024 = 464 b^2 - 16*b + 28 = 0 (b - 2)*(b - 14) = 0 b= 2 or b =14 xy = 2 or xy =14 etc etc

[@herbertklumpp2969]

I get the equation ( x+ y)^5 =1024 = x^5 +y^5 +5xy( x^3 +y^3) + 10x^2y^2( x +y) With x^3 +y^3 = ( x+y) ^3 -3xy(x+y) you get 1024= 464 + 5xy(64- 12 xy) + 10xy× 4. Therefore 28 = 16xy -x^2y^2 You get with t = x t = 14 or t=2 2= y(4-y) you get y= 2+ - sqrt(2) T=14 no real solution Y = 2+- i +sqrt(10)

[@is7728]

Wow the duration is so long

[@aines7958]

Excellent 👍👍👍 parfait 👍👋🌹🙏🇩🇿

[@SubasGhosh-bh7ns]

Awesome❤❤

[@biodiesel687]

I get the real solutions with a few iterations of Newton's method... Should there be a fifth solution? How would one find it?

[@cosmolbfu67]

x+y=a x^5+y^5=b fomular to find x.y 5a(xy)^2 + 5a^3(xy) + a^5 - b = 0

[@ТамараМошникова]

Уравнение от переменной просто решается по теореме Виета и далее простая система уравнений. Ход решения правильный, но не рациональный.

[@ylmazdagl9321]

G

[@yanfisher2639]

too much nonsense!