Are there any whole numbers that satisfy this equation?

[archaic]

Neither.
Get it into the form (a+...)(b+..)=...

you mean this $(a-x-y)(b+1)=y(y+1)-x(x-1)(x+1)$ ?

 

[haruspex]

you mean this $(a-x-y)(b+1)=y(y+1)-x(x-1)(x+1)$ ?

That's the idea, but you seem to have made some sign errors.
When you've fixed those, consider what order of picking values is likely to be helpful. If you plug in arbitrary values for a, b and one of x, y, you will have a quadratic or cubic to solve. What would be a better procedure?

 

[archaic]

That's the idea, but you seem to have made some sign errors.
When you've fixed those, consider what order of picking values is likely to be helpful. If you plug in arbitrary values for a, b and one of x, y, you will have a quadratic or cubic to solve. What would be a better procedure?

$$x^3-y^2+ab=a-b(x+y)\\x^3-y^2=a-ab-b(x+y)$$
$$k(a+X)(b+Y)+C=a-ab-b(x+y)\\kab+kaY+kbX+kXY+C=a-ab-b(x+y)\\k=-1,\,Y=-1,\,X=x+y,\,C=-(x+y)$$
$$x^3-y^2=-[a+(x+y)](b-1)-(x+y)\\x^3-y^2+x+y=-[a+(x+y)](b-1)\\x^3-y^2+x+y=(a+x+y)(1-b)$$
I have to rush to my bus now.

 

[archaic]

Two arrangements that stand out: when $b=1$ or $a=-x-y$.

 

[haruspex]

Two arrangements that stand out: when $b=1$ or $a=-x-y$.

a=-x-y is no use since all have to be positive, and b=1 is of limited value since it still leaves you with an awkward equation in x and y to solve.
There is a more interesting value of b, but note that if you increase b beyond 1 the left hand side goes negative. So you need to flip the sign both sides so that the factor on the left is b-1, not 1-b.
But you are still missing the point of the way I have encouraged you to rewrite the equation. The left hand side is now relatively easy to solve once you have plugged in values for x and y (as long as the RHS is positive).

 

[Charles Link]

Yes. Put $(a+x+y)$ on the left side. Since the OP is not responding, and @archaic has sort of taken over the function of the OP, let me show you what I got, and see if you agree. I get
$(a+x+y)(b-1)=y(y-1)-x(x^2+1)$.$\\$ Suggestion is to let $b=2$, $x=1$, let $y$ take on all kinds of values, and solve for $a$. This I think is what @haruspex has been looking for, and it is a very good way to get a bunch of solutions for this problem. Hopefully my algebra is accurate.

 

[haruspex]

Suggestion is to let b=2 , x=1 , let y take on all kinds of values, and solve for a .

Yes, that gives an infinite family of solutions. But there are plenty of others to be had by taking any value of x, then a sufficiently large value of y to make the RHS greater than x+y, then factorise that however you can.

 

[archaic]

Yes. Put $(a+x+y)$ on the left side. Since the OP is not responding, and @archaic has sort of taken over the function of the OP, let me show you what I got, and see if you agree. I get
$(a+x+y)(b-1)=y(y-1)-x(x^2+1)$.$\\$ Suggestion is to let $b=2$, $x=1$, let $y$ take on all kinds of values, and solve for $a$. This I think is what @haruspex has been looking for, and it is a very good way to get a bunch of solutions for this problem. Hopefully my algebra is accurate.

I was thinking of $b=2$ too, but I didn't go back to the equation so I thought it was $(b-1)$, not its opposite, and that for that $b$ I'd see $x+y$ cancel from the RHS and LHS, but it was $(1-b)$ :/

 

[haruspex]

I was thinking of $b=2$ too, but I didn't go back to the equation so I thought it was $(b-1)$, not its opposite, and that for that $b$ I'd see $x+y$ cancel from the RHS and LHS, but it was $(1-b)$ :/

To illustrate post #36, arbitrarily set x=5. That means y must be chosen to be at least 13, but that would only allow the b=2 solution we already have. So let's make y a bit larger, 15 say:
(a+20)(b-1)=80
We can factorise 80 in any way that makes one factor at least 21.. so 40x2, allowing a=20, b=3.

 

[archaic]

But there are plenty of others to be had by taking any value of x, then a sufficiently large value of y to make the RHS greater than x+y, then factorise that however you can.

$(a+x+y)(b-1)=-x^3+y^2-x-y$
I am not sure why you want the RHS to be greater than $x+y$. In the case of $b=2$, we can have $a$ to be negative in order to bring the sum back to the RHS. For $b$ different than $2$ I don't see why you want that.

 

[haruspex]

$(a+x+y)(b-1)=-x^3+y^2-x-y$
I am not sure why you want the RHS to be greater than $x+y$. In the case of $b=2$, we can have $a$ to be negative in order to bring the sum back to the RHS. For $b$ different than $2$ I don't see why you want that.

As clarified in post #14, all numbers have to be positive. The minimum of the LHS is x+y+1.

 

[SammyS]

As clarified in post #14, all numbers have to be positive. The minimum of the LHS is x+y+1.

Perhaps, it can be that $b = 1$. Then the LHS is zero.

 

[Charles Link]

Perhaps, it can be that b=1b=1b = 1. Then the LHS is zero.

See my post 23. Before doing it by @haruspex 's method, I did find a case where $b=1$, $x=1$ and $y=2$. $\\$ I see that $x=3$ and $y=6$ also works. If the whole numbers include zero, $x=0$ and $y= 1$ also works. $\\$ Looks like $a$ can be arbitrary=anything for these=it's obvious also by looking at the original expression in post 1.