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Are there any whole numbers that satisfy this equation?

  • Thread starter donglepuss
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  • #1
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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)
 

Answers and Replies

  • #2
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?
 
  • #3
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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.
 
  • #4
WWGD
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Try turning into a quadratic or cubic by selecting the right variable.
 
  • #5
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.
 
  • #6
DaveC426913
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And apparently it's debatable whether ##0## is a whole number.
Do you have a source that contests this?
 
  • #7
haruspex
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The a, b, ab terms suggest factorising how?
 
  • #9
DaveC426913
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  • #12
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.
 
  • #13
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.😏
 
  • #14
btw, i meant positive integers when i said whole numbers
 
  • #15
haruspex
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btw, i meant positive integers when i said whole numbers
Understood. Did you try my hint in post #7?
 
  • #16
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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##.
 
Last edited:
  • #18
DaveC426913
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Hi Dave:

I found a source that confirms your view.
Well, it wasn't "my" view :wink: - I checked a half dozen sources first, to verify the consensus that 0 is a whole number.
 
  • #19
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.
 
  • #20
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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}##
 
  • #21
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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  • #22
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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  • #23
Charles Link
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Suggest looking for a simple solution: Set ##a=b=1 ##. I think I found a very simple ##x ## and ##y ##.
 
  • #24
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.
 
  • #25
WWGD
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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.
 

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