$8x^4 + 8y^2 - 8xy^3 - 9x = 8y^4 + 8x^2 - 8yx^3 - 9y$ is equivalent to: $(x-y)(8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9)=0$ Consider the case when $x \not = y$ Then we have $8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ or $8x^4 + 16x^3y + 16x^2y^2 + 8xy^3 - 8x^2 - 8xy - 9x = 0$ From this equation substract the first from the system to obtain: $16x^3y + 16y^3x + 16x^2y^2 - 8x^2 - 8y^2 - 8xy = 0$ or $2xy(x^2 + y^2 + xy) = x^2 + y^2 + xy$ $x^2 + y^2 + xy$ is $0$ iff $x=y=0$ so we can divide by it to get: $y=\frac{1}{2x}$ Put this into the equation $8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ to see it's equivalent to: $x^6 = 1$ so $x=1$ and $y=\frac{1}{2}$ or $x=-1$ and $y=-\frac{1}{2}$. It's easy to check that the first pair is indeed a solution. We are left with the case $x=y$. It's not hard to verify that this condition yields 2 solutions. The final answer is: $(x, y) \in \{(0,0), (\frac{8}{9}, \frac{8}{9}), (1, \frac{1}{2})\}$

You didn't find all the solutions. Though I didn't analyse your solution carefuly but I think you lost some in the case $y=\frac{1}{2x}$.

I'm making stupid mistakes, as always. I hope it's correct right now.

That is right. Alternatively we can multiply the first equation by $y$, second equation by $x$, subtract one from another and factor. We get same conditions: $2xy-1=0$ or $x-y=0$. For the second condition, we can substitute $y=\frac{1}{2x}$ into the first equation, multiply it by $8x^2$, and get $8x^6-9x^3+1=0$, which yields $x=1$ or $\displaystyle\frac12$. So there are another two solutions: $(1,\frac12)$ and $(\frac12,1)$.

This problem comes from the Austrian-Polish MO 1985

Why do they all comes from the Austrian-Polish MO?I mean they can't find some new problem?

TomciO wrote: $8x^4 + 8y^2 - 8xy^3 - 9x = 8y^4 + 8x^2 - 8yx^3 - 9y$ is equivalent to: $(x-y)(8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9)=0$ Consider the case when $x \not = y$ Then we have $8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ or $8x^4 + 16x^3y + 16x^2y^2 + 8xy^3 - 8x^2 - 8xy - 9x = 0$ From this equation substract the first from the system to obtain: $16x^3y + 16y^3x + 16x^2y^2 - 8x^2 - 8y^2 - 8xy = 0$ or $2xy(x^2 + y^2 + xy) = x^2 + y^2 + xy$ $x^2 + y^2 + xy$ is $0$ iff $x=y=0$ so we can divide by it to get: $y=\frac{1}{2x}$ Put this into the equation $8x^3 + 16x^2y + 16xy^2 + 8y^3 - 8x - 8y - 9 = 0$ to see it's equivalent to: $x^6 = 1$ so $x=1$ and $y=\frac{1}{2}$ or $x=-1$ and $y=-\frac{1}{2}$. It's easy to check that the first pair is indeed a solution. We are left with the case $x=y$. It's not hard to verify that this condition yields 2 solutions. The final answer is: $(x, y) \in \{(0,0), (\frac{8}{9}, \frac{8}{9}), (1, \frac{1}{2})\}$ your solution is rigth, but I think 2nd solution is $(\frac{9}{8}, \frac{9}{8}$)$

The solution of his line too long!!!!!!!!!!. Solve the system of equations: $ \left\{\begin{aligned}x^{4}+y^{2}-xy^{3}-\frac{9}{8}x = 0 (1)\\ y^{4}+x^{2}-yx^{3}-\frac{9}{8}y=0 (2)\end{aligned}\right. $ My solution: $(1)*x^2$,$(2)*y^2$ then subtract side, hence $x^6-y^6-\frac{9}{8}x^3+\frac{9}{8}y^3=0$ equivalent to $(x^3-y^3)(x^3+y^3-\frac{9}{8})=0$ equivalent to $x=y or x^3+y^3=\frac{9}{8}$ with x=y (easy) $x^3+y^3=\frac{9}{8}$ (3) With $x^3+y^3=\frac{9}{8}$ then (1),(2):y then add side, we get $\frac{x^3+y^3}{xy}=\frac{9}{4}$ equivalent to $xy=1/2$, hence $x^3y^3=1/8$ (4) Since (3) and (4)............

1° case: If x,y≠0 and x/y isn't a cube root of unity: From the first equation, we have: x³ - y³ - 9/8 = -y²/x x³ + y³ - 9/8 = -y²/x + 2y³ Analogously, from the second equation, x³ + y³ - 9/8 = -x²/y + 2x³ Then, -y²/x + 2y³ = -x²/y + 2x³ (x³ - y³)/xy = 2(x³ - y³) xy = 1/2 Multiplying the first equation x² : x^6 - 9x³/8 + 1/8 = 0 x³ = 1 or 1/8 Analogously , y³ = 1 or 1/8 w.l.o.g. , x³=1 , y³=1/8 x=w^i , y=w^j/2 , i,j in {0,1,2} , w = cis(2π/3) = cos(2π/3) + i sin(2π/3) 3 l i+j , because xy=1/2 Checking the solutions, (x,y) = (1,1/2) , (1/2,1) , (w, w²/2) , (w²/2, w) , (w², w/2) , (w/2, w²) are solutions 2º case: If x,y≠0 and x/y is a cube root of unity: Then x³=y³. From the equations: y² - 9x/8 = 0 , x² - 9y/8 = 0 Easily can check (x,y) = (9/8, 9/8) , (9w/8, 9w²/8) , (9w²/8, 9w/8) are solutions in this case 3° case: If x=0 or y=0 In this case (x,y)=(0,0) is solution.