Solve …

…. y^3 = x^3 + 8x^2 – 6x + 8 for positive integers x and y.

There are many ways to approach this problem. One of the ways is to assume

y = x + a, for all a belongs to z

The problem can then be reduced to a quadratic expression in x. The discriminant then contains a fourth order expression in a. It gets tricky from there. So, a better approach to solve it is through reduction of available range of x and y. To start with, x and y could be in the range 0 to infinity.

y^3 = x^3 + 8x^2 – 6x + 8  ….. (1)

y^3 = (x^3 + 6x^2 + 12x + 8) + (2x^2 – 18x)

y^3 = (x^3 + 6x^2 + 12x + 8) + 2x(x – 9)

y^3 = (x+2)^3 + 2x(x – 9)….. (2)

y > 0 => y^3 >0

The equation reduces to a perfect cube when

2x(x-9) = 0 ….. (3)

x=0 or x=9

We have to now find if there could be any others values in range that can satisfy (1). To do this,

assume x = y + a where a is an integer

From (2),

y^3 – 2x(x – 9) = (x + 2)^3 

y^3 – 2y^2 + y(-4a + 18) + (18a – 2a^2) = (x+2)^3 …. (5)

If LHS of (5) were to be a perfect cube, then

((coefficient of y^2)/3) x ((coefficient of y)/3)  =  (constant)^2

This reduces to the quadratic equation

9a^2 – 77a – 18 = 0 

Neither of the roots of this quadratic equation is a positive integer.

It follows from this and (4) that only possible values of x are 0 and 9. For x = 0, y = 2 and for x = 9, y = 11. Since x and y are positive integers, in strict sense, x = 9 and y = 11 is only possible solution such that x and y are positive integers.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: