Are there any whole numbers that satisfy this equation?

[donglepuss]

Homework Statement

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

Relevant 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)

 

[SammyS]

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?

 

[gmax137]

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]

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

 

[LCKurtz]

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

 

[DaveC426913]

And apparently it's debatable whether $0$ is a whole number.

Do you have a source that contests this?

 

[haruspex]

The a, b, ab terms suggest factorising how?

 

[LCKurtz]

Do you have a source that contests this?

Not sure how serious a source this is, but:
https://www.lexico.com/en/definition/whole_number

 

[DaveC426913]

Not sure how serious a source this is, but:
https://www.lexico.com/en/definition/whole_number

Even if it were granted, it doesn't disqualify zero, (which is both fractionless and an integer).

 

[LCKurtz]

Even if it were granted, it doesn't disqualify zero, (which is both fractionless and an integer).

True enough, but some authors do.
http://mathworld.wolfram.com/WholeNumber.html

 

[haruspex]

True enough, but some authors do.
http://mathworld.wolfram.com/WholeNumber.html

Perhaps more relevantly, there are nontrivial positive solutions.

 

[DaveC426913]

Perhaps more relevantly, there are nontrivial positive solutions.

Yeah, we're just waiting for the OP to come back with some effort.

 

[haruspex]

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.

 

[donglepuss]

btw, i meant positive integers when i said whole numbers

 

[haruspex]

btw, i meant positive integers when i said whole numbers

Understood. Did you try my hint in post #7?

 

[archaic]

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 $x^2$.

 

[Buzz Bloom]

Do you have a source that contests this?

Hi Dave:

I found a source that confirms your view.
https://www.splashmath.com/math-vocabulary/number-sense/whole-numbers
Regards,
Buzz

 

[DaveC426913]

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]

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.

 

[archaic]

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]

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.)

 

[WWGD]

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.

 

[Charles Link]

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

 

[haruspex]

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.

 

[WWGD]

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.

But @archaic is not the OP, who has not shown any effort.

 

[haruspex]

But @archaic is not the OP, who has not shown any effort.

Yes, thanks, I should have referenced @donglepuss .

 

[Charles Link]

We are sort of in limbo=it's an interesting problem, but by PF rules, even if we have a whole set of solutions, we can't post them, but instead are waiting for the OP to see if he can find some solutions...

 

[haruspex]

We are sort of in limbo=it's an interesting problem, but by PF rules, even if we have a whole set of solutions, we can't post them, but instead are waiting for the OP to see if he can find some solutions...

True, it is frustrating. Happens all too often.
But maybe it is partly because different contributors are pushing in different directions. If you agree my method makes the problem easy, please post so.

 

[archaic]

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.)

do you mean get a as a function of b or the other way around?

 

[haruspex]

do you mean get a as a function of b or the other way around?

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

 

[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).