# Are there any whole numbers that satisfy this equation?

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## Homework Statement:

$x^3-y^2+ab=a-b(x+y)$ for any whole numbers $x,y,a,b$

## Homework Equations:

$x^3-y^2+ab=a-b(x+y)$
are there any whole numbers for x,y,a,b, that satisfy x^3-y^2+ab=a-b(x+y)

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SammyS
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Homework Statement: $x^3-y^2+ab=a-b(x+y)$ for any whole numbers $x,y,a,b$
Homework Equations: $x^3-y^2+ab=a-b(x+y)$

are there any whole numbers for x,y,a,b, that satisfy x^3-y^2+ab=a-b(x+y)
Before we can give help, you need to show some effort at getting a solution according to PF guidelines.

Once you do that, do you mean that $x,y,a,b$ actually need to be whole numbers, or is it that they need to be integers instead?

are there any whole numbers for x,y,a,b, that satisfy x^3-y^2+ab=a-b(x+y)
Yes. x=y=a=b=0.

WWGD
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Try turning into a quadratic or cubic by selecting the right variable.

LCKurtz
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$x=1,~y=1,~a=-4,~b=2$
Edit: Never mind. I guess whole numbers can't be negative. And apparently it's debatable whether $0$ is a whole number.

DaveC426913
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And apparently it's debatable whether $0$ is a whole number.
Do you have a source that contests this?

haruspex
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The a, b, ab terms suggest factorising how?

LCKurtz
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DaveC426913
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LCKurtz
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• DaveC426913
haruspex
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DaveC426913
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Perhaps more relevantly, there are nontrivial positive solutions.
Yeah, we're just waiting for the OP to come back with some effort.

haruspex
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Yeah, we're just waiting for the OP to come back with some effort.
I was concerned that the small hint I gave in post #7 might get lost amid the more arcane discussion. • DaveC426913
btw, i meant positive integers when i said whole numbers

• ryl3gol
haruspex
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btw, i meant positive integers when i said whole numbers
Understood. Did you try my hint in post #7?

I was concerned that the small hint I gave in post #7 might get lost amid the more arcane discussion. I don't see an apparent relation, what I could achieve for the moment is :
$$(x+y)(x-y-b)=-x^3+x^2+a-ab$$
by adding and subtracting $x^2$.

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DaveC426913
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Hi Dave:

I found a source that confirms your view.
Well, it wasn't "my" view - I checked a half dozen sources first, to verify the consensus that 0 is a whole number.

haruspex
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I don't see an apparent relation, what I could achieve for the moment is :
$$(x+y)(x-y-b)=-x^3+x^2+a-ab$$
by adding and subtracting $x^2$.
That's not what I have in mind. Think of a and b as the variables of a quadratic.

That's not what I have in mind. Think of a and b as the variables of a quadratic.
Yes, I have asked elsewhere and got the same answer. We can find that $y=\frac{b \pm \sqrt{b^2-4(-x^3-bx-ab+a)}}{2}=\frac{b \pm \sqrt{b^2+4x^3+4bx+4ab-4a}}{2}$

haruspex
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Yes, I have asked elsewhere and got the same answer. We can find that $y=\frac{b \pm \sqrt{b^2-4(-x^3-bx-ab+a)}}{2}=\frac{b \pm \sqrt{b^2+4x^3+4bx+4ab-4a}}{2}$
As I posted, think of a and b as the variables, x and y as constants, and factorise on that basis.
(It might be more obvious if you swap the roles of a, b with those of x, y.)

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WWGD
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Now we need to see if we can find conditions under which the discriminant is an integer ( necessary but not sufficient). Edit: Maybe we can do the same with other variables and put it together into a tighter condition.

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Suggest looking for a simple solution: Set $a=b=1$. I think I found a very simple $x$ and $y$.

haruspex
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Suggest looking for a simple solution: Set $a=b=1$. I think I found a very simple $x$ and $y$.
As I keep trying to get @archaic to do, it is much easier to find solutions if you flip it around, thinking of a and b as the dependent variables. See posts #7 and #21.

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