$y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)} $ $x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$

UzbekMathematician wrote: $y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)} $ $x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$ Good solution !

UzbekMathematician wrote: $y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)} $ $x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$ That's how I solved it

UzbekMathematician wrote: $y^2-x^2y-x^3=0 \implies $ $\ldots$ Small typo. It needs to be $y^2+x^2y-x^3=0$.

nevermind

JANMATH111 wrote: Solved it during the contest: x=0 <==> y=0 now x and y are not 0. Let gcd(x, y)=d Then x=dk and y=dl and gcd(k, l)=1 then plugging in that stuff gives k^2ld+l^2=k^3d after division by d^2 Now we see that d|l^2 so l^2=nd. plugging that in again and dividing by d gives k^2l+n=k^3 now we see that k^2|n|l^2 so k and l are not coprime after all... THat means k=l=1 but that is also not a solution so x=y=0. $(-4,-8)$ is also a solution.

This is my solution on the contest, similiar with @UzbekMathematician We can rewrite the equation as $y^2+x^2y-x^3=0$ Since the coefficient of $y^2$ on $LHS$ is $1$ which is different from $0$, we can solve this equation as a quadratic equation with $y$ as variable. For $y$ to be an integer, it's neccessary for the discriminant to be a perfect square, so $x^4+4x^3=x^2x(x+4)=m^2$ where $m$ is an integer. This happens if and only if $x(x+4)$ is a perfect square, so lets write $x(x+4)=l^2$, where $l$ is an integer. So $x^2+4x-l^2=0$. For the same reason, we solve this equation, as a quadratic equation, where $x$ is a variable. For $x$ to be an integer it's neccessary for the discrimant to be a perfect square. So we have $16+4l^2=k^2$, where $k$ is an integer. So, $16=(k-2l)(k+2l)$ By checking all cases: $16=16\cdot 1, 16=8\cdot 2, 16=4\cdot 4, 16=2\cdot 8, 16=1\cdot 16$ And the cases where the factors are negative, e.g $16=(-16)\cdot (-1)$ ( I checked all the cases one by one on the contest ), and from all the solutions I got that $l=0$, so $x(x+4)=0$, so $x=0$ or $x=-4$. By substituting on the original equation we get that the only solutions are $(0,0)$ and $(-4,-8)$.

$y(x^2+y)=x^3$. Thus $x=ky$, where $k\in{\mathbb{Q}}$ Simplifying gives $y=y^2(k^3-k^2)$ which gives $(0,0)$ as solution Otherwise we have $y(k^3-k^2)=1$ so $\frac{1}{k^3-k^2} \in{\mathbb{Z}}$ Let $k=\frac{p}{q}$ where $(p,q)=1$ Then $\frac{q^3}{p^2-p^2q} \in{\mathbb{Z}}$ $(p^2(p-q))\mid{q^3}$ but $(p,q)=1$ Case 1: $p=q-1$ $\frac{(1+p)^3}{-p^2}$ integer so $p=1, q=2$ Case 2: $p=q+1$ Note implies $p,q$ both negative so solution covered in case $1$ Thus $y=2x$ giving $(-4,-8)$ as well Thus we've exhaused our solution set $\blacksquare$

If $x=0$ so $y=0$. Now $x,y$ different of zero. Divide the equation $x^2$ so $y+\frac{y^2}{x^2}=x$. Thus $y=kx$. So $xk+k^2=x$. Then $x=-\frac{k^2}{k-1}$ with $k$ different of zero. So the unique solution is $k=2$. Then $(x,y)=(-4,-8)$ and $(0,0)$.

Hello. Obvious solution is $x=y=0$, so we will find another solutions. We can factorize this $x^2y+y^2-x^3=0$, now we will divide both sides with $x^2$ and we will get: $\frac{y^2}{x^2}+y-x=0$. We will solve this equation using quadratic formula for $y$ and using that $a=\frac{1}{x^2}, b=1, c=-x$, so: $y=\frac{-1 \pm \sqrt{1+\frac{4}{x}}}{\frac{2}{x^2}}$ Since we only have one solution for $y$, it means that discriminant has to be $0$, so: $\sqrt{1+\frac{4}{x}}=0$ and finally $x=-4$ and $y=-8$. So, solutions are: $(x,y)=(0,0);(-4,-8)$

$x^2y+y^2=x^3$ (1) ($x=0$ and $y=0$ is true and let's say that $x$,$y$ isn't 0) Since $x$ and $y$ are integers we know that $x^2$|$y^2$ which means $x$|$y$ Let's say that $y$=$ax$ and put this in (1) We get $ax^3$+$a^2x^2$=$x^3$ and since $x$ isn't 0 divide both sides by $x^2$ We get $ax+a^2=x$ now $a$|$x$ and let's say $x$=$ab$ put this $x$ and we get: $a^2b+a^2=ab$ now divide both sides by $a$ and we get: $ab+a=b$ and from here $a$|$b$ and $b$|$a$ which means that $a=b$(2) or $a=-b$(3) From (2) we get $(x;y)=(0;0)$ From (3) we get $-a^2+a=-a$ which is same as $a^2=2a$ so $a=2$(different from0) and from this we get $(x,y)=(-4,-8)$ $ANSWER:(x,y)=(0,0),(-4,-8)$