Are there any whole numbers that satisfy this equation?

[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.