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Anonymous Sat 27 Jun 07:31:37 2020 No.11841783 View Reply Original Report

Quoted By: >>11842788 >>11843039

x^3-y^2+ab=a-b(x+y)

are there any positive integers x,y,a,b that satisfy the equation when x, y, a, and b are all different numbers?

and furthermore, if there are, is there an infinite number of solutions?

Anonymous

Anonymous Sat 27 Jun 2020 10:37:35 No.11842076 Report

Quoted By:

x^3-y^2+ab = a-b(x+y)
x^3-y^2 = a-b(x+y+a)
x^3-y^2-a = -b(x+y+a)
y^2+a-x^3 = b(x+y+a)
(y^2+a-x^3)/(x+y+a) = b

So you need to find integer solutions to (y^2+a-x^3)/(x+y+a)

Anonymous

Anonymous Sat 27 Jun 2020 16:21:20 No.11842788 Report

Quoted By:

>>11841783
a = 42
b = 1
x = 1
y = -1

Anonymous

Anonymous Sat 27 Jun 2020 18:13:32 No.11843039 Report

Quoted By: >>11843072

>>11841783
there are a bunch of solutions
x : 1, y : 4, a : 5, b : 2
x : 1, y : 6, a : 7, b : 3
x : 1, y : 7, a : 2, b : 5
x : 1, y : 8, a : 9, b : 4
x : 1, y : 9, a : 4, b : 6
x : 3, y : 6, a : 2, b : 1
x : 3, y : 6, a : 4, b : 1
x : 3, y : 6, a : 5, b : 1
x : 3, y : 6, a : 7, b : 1
x : 3, y : 6, a : 8, b : 1
x : 3, y : 6, a : 9, b : 1
x : 3, y : 9, a : 2, b : 4
don't know about infinite, but almost surely yes

Anonymous

Anonymous Sat 27 Jun 2020 18:23:13 No.11843072 Report

Quoted By:

>>11843039
actually, infinite is super easy
just take $x = 1, b = 2$ , $y > 3$ arbitrary and $a = y^2 - 2 y - 3$

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