Can You Solve This Diophantine Equation? | Math Olympiad Algebra

[@abcekkdo3749]

(x,y)=(6,-3) (-3,6) (-12,-3) (-3,-12)

[@RealQinnMalloryu4]

({xy+xy ➖ } +{3+3 ➖ }({ x+ x ➖ } )/{9x^2+9y^2}= ({xy^4+6+ {x^2+y^2})/18xy^4 ={6xy^4 +xy^4}/18xy^4= 6xy^8/18xy^4 = 3xy^2 (xy ➖ 3xy+2).

[@walterwen2975]

Math Olympiad algebra: [xy + 3(x + y)]²/[(x + 3)² + (y + 3)²] = 1, x , y ϵZ ≠ - 3
Let: u = x + 3, v = y + 3; x + y = u + v - 6, xy = (u - 3)(v - 3), u, v ϵZ
[xy + 3(x + y)]²/[(x + 3)² + (y + 3)²] = [(u - 3)(v - 3) + 3(u + v - 6)]²/(u² + v²) = 1
[uv - 3(u + v) + 9 + 3(u + v) - 18]² = (uv - 9)² = u²v² - 18uv + 81 = u² + v²
(u² + 2uv + v²) - (u²v² - 16uv + 64) - 17 = 0, (u + v)² - (uv - 8)² = 17
(u + v)² > (uv - 8)² > 17; 17 is a prime number
[(u + v) + (uv - 8)][(u + v) - (uv - 8)] = (1)(17) = (- 17)(- 1); u, v integers
(u + v) + (uv - 8) = 1, (u + v) - (uv - 8) = 17; u + v = 9, uv - 8 = - 8, uv = 0
u + v = x + y + 6 = 9, x + y = 3; uv = (x + 3)(y + 3) = 0, x = - 3 or y = - 3
x = - 3; y = 3 - x = 6 or y = - 3; x = 3 - y = 6
(u + v) + (uv - 8) = - 17, (u + v) - (uv - 8) = - 1; u + v = - 9, uv - 8 = - 8, uv = 0
u + v = x + y + 6 = - 9, x + y = - 15; x = - 3 or y = - 3
x = - 3; y = - 15 - x = - 12 or y = - 3; x = - 15 - y = - 12
x = - 3, y = 6; x = 6, y = - 3; x = - 3, y = - 12 or x = - 12, y = - 3
Answer check:
[xy + 3(x + y)]²/[(x + 3)² + (y + 3)²] = 1
x = - 3, y = 6: [- 18 + 3(3)]²/[0 + (6 + 3)²] = (- 9)²/9² = 1; Confirmed
x = 6, y = - 3: [- 18 + 3(3)]²/[(6 - 3)² + 0] = (- 9)²/9² = 1; Confirmed
x = - 3, y = - 12: [36 + 3(- 15)]²/[0 + (- 9)²] = (- 9)²/(- 9)² = 1; Confirmed
x = - 12, y = - 3: [36 + 3(- 15)]²/[(- 9)² + 0] = (- 9)²/(- 9)² = 1; Confirmed
Final answer:
x = - 3, y = 6; x = 6, y = - 3; x = - 3, y = - 12 or x = - 12, y = - 3