Cracking an Ultimate System of Equations 🌟 | Math Olympiad

[@9허공]

another approach
let a = √x, b = √(x + 32), c = y - 1 then a^2 - b^2 = -32 --- (1)
√x + √(x + 32) = (2 - y)^5 => a + b = - (c - 1)^5 --- (2) since a + b >= 0 , c a - b = (c + 1)^5 --- (3)
(2)*(3) => (a + b)*(a - b) = a^2 - b^2 = -32 = -2^5 = -(c - 1)^5*(c + 1)^5 = -((c - 1)*(c + 1))^5
(c - 1)(c + 1) = c^2 - 1 = 2 => c^2 = 3, since c 2a = - (c - 1)^5 + (c + 1)^5 = 2*(5c^4 + 10c^2 + 1) =2*(5*9 + 10*3 + 1) = 2*76
=> a = 76 => x = 76^2 = 5776 , y = c + 1 = 1 - √3, answer (x, y) = (5776, 1 - √3)

[@abcekkdo3749]

（x,y)=(5776,1-√3)

[@mohammedsaysrashid3587]

Thank you for sharing Sir🙏....(x,y) =( 5776,-0,731)

[@davidshen5916]

Sqrt（X）+Sqrt（X+32）=（2-Y）^5， Sqrt（X）- Sqrt（X+32）=Y^5， 1️⃣*2️⃣ X-（X+32）=（2Y-Y^2）^5， （Y^2-2Y）^5-2^32=0， Let A=（Y^2-2Y）/2， A^5-1=0， （A-1）（A^4+A^3+A^2+A+1）=0， A=1， Y^2-2Y=2， Y=1+-Sqrt（3）， Y

[@ZhilinChen-my7tp]

Y=1-√3

[@ZhilinChen-my7tp]

（2-y）y=-2，2y-y^2+2=0，y=1+√3，1-√3，2-1-√3≤0 is rejected，so，y=1-√3。