The only integer solutions of the Diophantine equation $(1) \;\; x^3(y + 1) + y^3(x + 1) = 19$ are $(x,y)= (-20,-1), (-1,-20), (1,2) ,(2,1)$. Proof: We may WLOG assume (since $x$ and $y$ are symmetric in the LHS of equation (1)) $x \leq 0$. Clearly $x,y \neq 0$ by equation (1). If $x < 0 < y$, then $x^3(y + 1) < 0$ and $y^3(x + 1) \leq 0$, implying the LHS of equation (1) is negative. This contradiction implies $x$ and $y$ have the same sign. Hence we have the following two cases to consider: Case 1: $x \leq y < 0$. Hence $(u,v) = (-x,-y)$ give us $u \geq v > 0$ and (according to equation (1)) $(2) \;\; u^3(v - 1) + v^3(u - 1) = 19$, which (since $u \leq v > 0$) yields $19 \geq v^3(v - 1) + v^3(v - 1)$, which means $(3) \;\; v^3(v - 1) \leq 9$. Hence $v \leq 2$ by inequality (3). If $v=1$, then $u=20$ by equation (2). If $v=2$, then $u(u^2 + 8) = 27$ by equation (2), implying $u \not \in \mathbb{N}$. Consequently the only solution of equation (2) is $(u,v) = (20,1)$. Case 2: $0 < x \leq y$. Then by equation (1) $19 \geq x^3(x + 1) + x^3(x + 1)$, yielding $(4) \;\; x^3(x + 1) \leq 9$. Therefore $x=1$ by inequality (4), which inserted in equation (1) result in $(5) \;\; y(2y^2 + 1) = 18$. Hence $y=2$ by equation (4), which give us the solution $(x,y)=(1,2)$ of equation (1). Summa summarum, the only integer solutions of equation (1) with $x \leq y$ are $(x,y)=(-20,-1)$ and $(x,y)=(1,2)$. q.e.d.

The good problem

Generalizing... XY^n+YX^n+X^n+Y^n=19

$x^3(y+1)+y^3(x+1)=$ $xy(x^2+y^2)+(x^3+y^3)=$ $xy((x+y)^2-2xy)+(x+y)((x+y)^2-3xy)=19$ Let $a=x+y$ and $b=xy$. Then the equation becomes $$2b^2-ba^2+3ab-a^3+19=0$$The discriminant over $b$ is $$\Delta=a^4+2a^3+9a^2-152$$Since $b$ is an integer, $\Delta=k^2$ for $k$ is an integer. First notice that $\Delta<0$ for $-3 \leq x \leq 2$. It is easy to see that $(a^2+a+5)>\Delta$ and for $|x|>4$, $(a^2+a)<\Delta$. So $\Delta$ must be equal to $(a^2+a+4)$, $(a^2+a+3)$, $(a^2+a+2)$ or $(a^2+a+1)$. Basic algebra gives us $\Delta= a^4+2a^3+9a^2-152=(a^2+a+4)^2$ $\implies$ $a=-21$ and $b=20$. For $a=\{-4,3,4\}$, $a=3$ gives us $\Delta=64$ and $b= \pm 2$. $(a,b)=(-21,20)\implies(x,y)=(-20,-1)$ and $(a,b)=(3,2)\implies(x,y)=(2,1)$. $b=-2$ clearly doesn’t work. So the final answer is $$(x,y)=\{(2,1),(1,2),(-20,-1),(-1,-20)\} \; \blacksquare$$

I have solve this using inequalities

How? Can you post it?

A wasted opportunity for a hard problem. The idea is to use $s=x+y$, $p=xy$, but because $x+1$ and $x^3$ are of odd degree, simple bounding works, undermining a N4: Case $1:$ $x=-2$: $-8(y+1)-y^3=19\implies y^3+8y+27=0$, no integer solutions Case $2:$ $x=-1$: $-(y+1)=19\implies y=-20$, so we found the solutions $(x,y)=(-1,-20)$ and $(x,y)=(-20,-1)$, because we have symmetry between $x$ and $y$, so if $(x,y)$ is a solution, then $(y,x)$ is a solution too. Case $3:$ $x=0$: $y^3=19\implies$ no integer solutions Case $4:$ $x=1$: $(y+1)+2y^3=19\implies 2y^3+y-18=0\implies (y-2)(2y^2+4y+7)=0\implies y=2$ as $2y^2+4y+7=2(y+1)^2+3>0$ and we've found the solutions $(x,y)=(1,2),(2,1)$. Case $5$: $x=2$: $x^3(y+1)+y^3(x+1)=8y+8+3y^3=19\implies 3y^3-3+8y-8=0\implies (y-1)(3y^2+3y+11)\implies y=1$ as $3y^2+3y+11>0$, so we've reached the same solution as in Case $4$. Now onto the more general cases Case $6$: $|x|>2,|y|>2$, $xy<0$: $x^3(y+1)+y^3(x+1)\leq-54-54=-108<19$, so no solutions here. Case $7$: $|x|>2,|y|>2$, $xy>0$: $x^3(y+1)+y^3(x+1)\geq 24+24>19$, so no solutions here either. Finally, all the solutions are $\boxed{(x,y)=(1,2),(2,1),(-1,-20),(-20,-1)}$

mrdriller wrote: $x^3(y+1)+y^3(x+1)=$ $xy(x^2+y^2)+(x^3+y^3)=$ $xy((x+y)^2-2xy)+(x+y)((x+y)^2-3xy)=19$ Let $a=x+y$ and $b=xy$. Then the equation becomes $$2b^2-ba^2+3ab-a^3+19=0$$The discriminant over $b$ is $$\Delta=a^4+2a^3+9a^2-152$$Since $b$ is an integer, $\Delta=k^2$ for $k$ is an integer. First notice that $\Delta<0$ for $-3 \leq x \leq 2$. It is easy to see that $(a^2+a+5)>\Delta$ and for $|x|>4$, $(a^2+a)<\Delta$. So $\Delta$ must be equal to $(a^2+a+4)$, $(a^2+a+3)$, $(a^2+a+2)$ or $(a^2+a+1)$. Basic algebra gives us $\Delta= a^4+2a^3+9a^2-152=(a^2+a+4)^2$ $\implies$ $a=-21$ and $b=20$. For $a=\{-4,3,4\}$, $a=3$ gives us $\Delta=64$ and $b= \pm 2$. $(a,b)=(-21,20)\implies(x,y)=(-20,-1)$ and $(a,b)=(3,2)\implies(x,y)=(2,1)$. $b=-2$ clearly doesn’t work. So the final answer is $$(x,y)=\{(2,1),(1,2),(-20,-1),(-1,-20)\} \; \blacksquare$$[/qu Wow it was exacly the same solution as I did while I was doing on my own . ( a and b variables were same as well.

$x^3(y+1)+y^3(x+1)=$ $xy(x^2+y^2)+(x^3+y^3)=$ $xy((x+y)^2-2xy)+(x+y)((x+y)^2-3xy)=19$ Let $a=x+y$ and $b=xy$. Then the equation becomes $$2b^2-ba^2+3ab-a^3+19=0$$The discriminant over $b$ is $$\Delta=a^4+2a^3+9a^2-152$$Since $b$ is an integer, $\Delta=k^2$ for $k$ is an integer. First notice that $\Delta<0$ for $-3 \leq x \leq 2$. It is easy to see that $(a^2+a+5)>\Delta$ and for $|x|>4$, $(a^2+a)<\Delta$. So $\Delta$ must be equal to $(a^2+a+4)^2$, $(a^2+a+3)^2$, $(a^2+a+2)^2$ or $(a^2+a+1)^2$. Basic algebra gives us $\Delta= a^4+2a^3+9a^2-152=(a^2+a+4)^2$ $\implies$ $a=-21$ and $b=20$. For $a=\{-4,3,4\}$, $a=3$ gives us $\Delta=64$ and $b= \pm 2$. $(a,b)=(-21,20)\implies(x,y)=(-20,-1)$ and $(a,b)=(3,2)\implies(x,y)=(2,1)$. $b=-2$ clearly doesn’t work. So the final answer is $$(x,y)=\{(2,1),(1,2),(-20,-1),(-1,-20)\} \; \blacksquare$$

My original post got deleted so I am posting again

Steve12345 wrote: Find all integers $x,y$ such that $x^3(y+1)+y^3(x+1)=19$. Proposed by Bulgaria İf $y$$=$$0$ then there is no solution $y=-1$ $x=-20$,$y=-20$ $x=-1$ $x,y$$\neq$$-1,0$then $x$$\ge$0 $y$$\le$$0$it is impossible $x$$\le$0 $y$$\ge$$0$it is impossible Then $x,y$ must be both positive or both negative $x,y$$\ge$0 it is known that $x,y$$\le$$2$ if we look at the cases $x=1$ $y=2$ and $x=2$ $y=1$ $x,y$$\le$0 $x=-x_1$ and $y=-y_1$ $x_1(y_1-1)$$+$$y_1(x_1-1)$$=$$19$ İt is know that $x_1,y_1$$\neq$1,0 $x_1,y_1$$\le$$2$ if we look at the cases, we will see that there is no solution here $(x,y)$$=$${(1,2);(2,1)(-1,-20)(-20,-1)}$

Marinchoo wrote: A wasted opportunity for a hard problem. The idea is to use $s=x+y$, $p=xy$, but because $x+1$ and $x^3$ are of odd degree, simple bounding works, undermining a N4: Case $1:$ $x=-2$: $-8(y+1)-y^3=19\implies y^3+8y+27=0$, no integer solutions Case $2:$ $x=-1$: $-(y+1)=19\implies y=-20$, so we found the solutions $(x,y)=(-1,-20)$ and $(x,y)=(-20,-1)$, because we have symmetry between $x$ and $y$, so if $(x,y)$ is a solution, then $(y,x)$ is a solution too. Case $3:$ $x=0$: $y^3=19\implies$ no integer solutions Case $4:$ $x=1$: $(y+1)+2y^3=19\implies 2y^3+y-18=0\implies (y-2)(2y^2+4y+7)=0\implies y=2$ as $2y^2+4y+7=2(y+1)^2+3>0$ and we've found the solutions $(x,y)=(1,2),(2,1)$. Case $5$: $x=2$: $x^3(y+1)+y^3(x+1)=8y+8+3y^3=19\implies 3y^3-3+8y-8=0\implies (y-1)(3y^2+3y+11)\implies y=1$ as $3y^2+3y+11>0$, so we've reached the same solution as in Case $4$. Now onto the more general cases Case $6$: $|x|>2,|y|>2$, $xy<0$: $x^3(y+1)+y^3(x+1)\leq-54-54=-108<19$, so no solutions here. Case $7$: $|x|>2,|y|>2$, $xy>0$: $x^3(y+1)+y^3(x+1)\geq 24+24>19$, so no solutions here either. Finally, all the solutions are $\boxed{(x,y)=(1,2),(2,1),(-1,-20),(-20,-1)}$ Why do you say 'x+1 and x^3 are of odd degree' would motivate bounding?