Question

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Answer 1

By closure property of multiplication and addition of integers,

If
`x + (1)/(x)` is an integer

∴
`\left ( x + (1)/(x) \right) ^3 = x^3 + (1)/(x^3) +3\cdot \left (x + (1)/(x) \right )` is an integer

From which we have;

`x^3 + (1)/(x^3)` is an integer

Explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

`x + (1)/(x)`

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

`\left ( x + (1)/(x) \right) ^3 = x^3 + (1)/(x^3) +3\cdot x + (3)/(x)`

By simplification of the cube of the given integer expressions, we have;

`\left ( x + (1)/(x) \right) ^3 = x^3 + (1)/(x^3) +3\cdot \left (x + (1)/(x) \right )`

Therefore, we have;

`\left ( x + (1)/(x) \right) ^3 - 3\cdot \left (x + (1)/(x) \right )= x^3 + (1)/(x^3)`

By rearranging, we get;

`x^3 + (1)/(x^3) = \left ( x + (1)/(x) \right) ^3 - 3\cdot \left (x + (1)/(x) \right )`

Given that
`x + (1)/(x)` is an integer, from the closure property, the product of two integers is always an integer, we have;

`\left ( x + (1)/(x) \right) ^3` is an integer and
`3\cdot \left (x + (1)/(x) \right )` is also an integer

Similarly the sum of two integers is always an integer, we have;

`\left ( x + (1)/(x) \right) ^3 + \left(- 3\cdot \left (x + (1)/(x) \right ) \right )` is an integer

`\therefore x^3 + (1)/(x^3) = \left ( x + (1)/(x) \right) ^3 - 3\cdot \left (x + (1)/(x) \right )= \left ( x + (1)/(x) \right) ^3 + \left(- 3\cdot \left (x + (1)/(x) \right ) \right )` is an integer

From which we have;

`x^3 + (1)/(x^3)` is an integer.