Peter wrote: Find all pairs of integers $ (x, y)$ satisfying the equality \[ y(x^{2} + 36) + x(y^{2} - 36) + y^{2}(y - 12) = 0. \] $ yx^2+x(y^2-36)+y^2(y-12)+36y$ $ \Delta=(y^2-36)^2-4y(y^3-12y^2+36y)$ $ =-3y^4+48y^3-216y^2+36^2$ $ =-3(y^4-16y^3+72y^2-432)$ Easy to find out $ y$ to $ \Delta\geq 0$ Try to find out $ y$ and solve with x. Note: Are there any method? It is quite hard to carculas.

: I don't think you proved anything here?

pavel kozlov wrote: Consider our equation as quadratic respectively $ x$: $ yx^{2} + (y^{2} - 36)x + y(y - 6)^{2} = 0$. It's discriminant is the next: $ D = (y^{2} - 36)^{2} - 4y^{2}(y - 6)^{2} = (y - 6)^{2}((y + 6)^{2} - 4y^{2}) = - 3(y - 6)^{2}(y^{2} - 4y - 12) = - 3(y - 6)^{3}(y + 2)$. But $ D\geq 0$, hence $ - 2\leq y\leq 6$. 1) $ y = - 2$: $ - 2x^{2} - 32x - 128 = 0$ $ x^{2} + 16x + 64 = 0$ $ x = - 8$; 2) $ y = - 1$: $ D = 21*7^{2}$ - is not perfect square; 3) $ y = 0$: $ - 36x = 0$ $ x = 0$; 4) $ y = 1$: $ D = 3^{2}*5^{2}*5$ - is not perfect square; 5) $ y = 2$: $ D = 3*4^{4}$ - is not perfect square; 6) $ y = 3$: $ D = 3^{4}*5$ - is not perfect square; 7) $ y = 4$: $ 4x^{2} - 20x + 16 = 0$ $ x^{2} - 5x + 4 = 0$ $ x = 1,4$; 8) $ y = 5$: $ D = 21$ - is not perfect square; 9) $ y = 6$: $ 6x^{2} = 0$ $ x = 0$. Answer: $ ( - 8, - 2); (0,0); (0,6); (1,4); (4,4)$.

the equation mod x yields y = 0, 6 mod x. y = 0 implies y = kx which leads to the cubic $(k^3+k^2+k)x^3+(-12k^3)x^2+(36(k-1))x = 0$. It can be seen from the quadratic formula that x = $\frac{6k^2\pm6\sqrt{k}}{k^3+k^2+k} \rightarrow k = 1, k =4$ (note that at k = 9 and beyond the denominator is too big to make x an integer) which implies x=y=4, x = 1, y = 4, and x=y=0 are solutions. For the case y = kx+6 we obtain the cubic $x^2(x(k^3+k^3+k)+6k^2+12k+6) = 0 \rightarrow x = \frac{-6(k+1)^2}{k^3+k^2+k} \rightarrow k = 1, -2 \rightarrow x = -8, 1, 0 \rightarrow y = -2, 4, 6$ checking yields that x=-8, y = -2 and x = 0, y = 6 are also solutions.