The only solutions are $(0,0), (1,1), (2,2)$, which obviously work. Taking the equation modulo $xy$, we see that $xy$ must divide $x + y$. Since $x,y$ are nonnegative, this also implies that $xy \le x +y$, so $(x-1)(y-1) \le 1$. If $x,y$ were both greater than $1$, we would have $x-1$ and $y-1$ are positive integers multiplying to a number at most $1$, meaning that $x = y = 2$. If one of $x,y$ were $0$, then clearly the other one is zero also. If $x=1$, then $y\mid y + 1\implies y =1$. If $y = 1$, then $x\mid x + 1\implies x =1$. We have exhausted all cases, so we are done .

Pretty much the same as above; writing for storage. Case 1: $xy = 0$. In this case the other is also visibly zero, so we're done. Case 2: $xy \ne 0$. Observe that we get $x + y = xy(1 + 2y - x)$. In particular, $xy \mid x+y \implies xy \le x+y \implies (x-1)(y-1) \le 1$. Now if $(x-1)(y-1) = 0$, we have $x = 1$ or $y = 1$, and in both cases one can check that the other is $1$ as well. Otherwise, $x - 1 = y-1 = 1 (\text{ It can't be } -1 \text{ because that would mean } x y = 0)$. In other words, $x = y = 2$. Since all the solutions $(0, 0), (1, 1), (2, 2)$ work, we're done with the problem and these are the solutions. $\square$

A bit different from the above but the same in essentials I guess. If one of $x,y$ are 0 so is the other. So, assume $x,y\neq 0$ in what follows. Note that the given Diophantine equation rewrites to, \[x^3y+x+y=xy(2y+1)\]Thus, $x \mid x^3y+x+y \implies x\mid y$ and $y \mid x^3y+x+y \implies y\mid x$. Thus, we must have $x=y$. Now, plugging this in we obtain, \begin{align*} x^4+2x &= x^2+2x^3\\ x^3-2x^2-x+2 &=0\\ (x^2-1)(x-2) &=0 \end{align*}Thus, $x^2 =1 \implies x=1$ or $x=2$. Thus, the only solutions to the given equation are $(0,0) ; (1,1)$ and $(2,2)$ as desired.

https://artofproblemsolving.com/community/c6h2690610p23353742 This problem is extremely similliar to what I posted 3 years before.. quite interesting....

If $xy \neq 0 \Longrightarrow x \mid y$ and $y \mid x$ so, $x = y \Longrightarrow$ $x^4 + 2x = x^2 + 2x^3 \Longrightarrow$ $x^2(x^2 - 1) = 2x(x^2 - 1)$ thus: $\bullet$ $x^2 - 1 = 0$ $\Longrightarrow x = 1$ Hence, $(x, y) = (1, 1)$ $\bullet$ $x^2 - 1 \neq 0$ $\Longrightarrow x^2 = 2x$ $\Longrightarrow x = 2$ Hence, $(x, y) = (2, 2)$ If $xy = 0$ obviously $(x, y) = (0, 0)$ Therefore, $\boxed{(x, y) = (0, 0); (1, 1); (2, 2)}$

Claim: $(1,1)$, $(2,2)$, and $(0,0)$ are the only possible solutions. Proof: We see that if either of $x$ or $y$ is $0$, the other must be $0$ too. So $(0,0)$ is clearly a solution. Now let $(x,y)\in \mathbb{Z^+}$. We have $y\equiv 0\pmod x$ and $x\equiv 0\pmod y$. This gives $x=y$. Plugging $x=y$ in the original equation, we get $(1,1)$ and $(2,2)$ are the only other solutions.

Also here: https://artofproblemsolving.com/community/c6h423490p2395322

When either $x=0$ or $y=0$ it is easy to verify $x=y=0$ is solution to the problem.Then assume $x$ and $y$ are not equal to zero. We have $x^{3}y+x+y=xy(2y+1)$. Because $RHS$ is divisible by $x$ that means $LHS$ is divisible by $x$.From that we get $x|y$. Similarly we get $y|x$, from that we get $x=y$. Plugging $y$ to $x$ gives us $x^4+2x=2x^3+x^2$ $x^3-2x^2-x+2=0$ $(x^2-1)(x-2)=0$ That means $x^2=1$ and $x=2$. Then we get 3 solutions $(x,y)=(0,0);(1,1);(2,2)$

$(0, 0)$ works, so we can assume that $x, y \neq 0$ and we can now take mods. Take the equation $\pmod{x}, \pmod{y}$, to get that $x \equiv 0 \pmod{y}$ and $y \equiv 0 \pmod{x}$, implying $x=y$. Then $x^4+2x=x^2+2x^3 \implies x^3+2=x+2x^2$. The roots of this are $-1, 1, 2$ so our only solutions are $\boxed{(0, 0), (1, 1), (2, 2)}$

FAA2533 wrote: Find all non-negative integers $x, y$ such that\[x^3y+x+y=xy+2xy^2\] wattesigma take mod x then either one of x and y is 0 in which this is easy or x+y=0 mod xy you can bound this easily also ig x=y so u are done