An interesting relation

I came across this problem while reading Euler’s Elements of Algebra, and decided to solve it in my own way. The question is as thus:

Find two numbers, whose sum is equal to their product, which in turn is equal to the difference of their squares.

Let the two numbers be x and y, such that x >= y

Then, x + y = xy = x² – y²

The one obvious solution is when both x and y are equal to zero, so we ignore that and look for any other solutions it may have.

Let’s evaluate the following equality:

x + y = x² – y²

x + y = (x + y)(x – y) ----(1)

Now I state that x + y cannot be equal to zero.

If it were, then x + y = xy = 0 so either x or y is zero.

But since x + y = 0, if any one of them is equal to zero, the other has to be zero as well.

And we already overlooked the case where x = y = 0

So we can cancel out (x + y) from equation (1) to get:

1 = x – y

which means that: x = y + 1 ----(2)

Now let us evaluate another equality:

x + y = xy

Substituting x from (2),

y + y + 1 = (y + 1)y

2y + 1 = y² + y

y² = y + 1 ----(3)

Look familiar? Let’s solve it.

y² – y – 1 = 0

Using the quadratic formula,

y = [1 ± √(1 + 4)]/2

= [1 ± √5]/2

Note: Yes, this is equal to the golden ratio. (Look it up on wikipedia if you don’t know it.)

And x = y + 1 = [3 ± √5]/2

So the two numbers could either be:

[1 + √5]/2 and [3 + √5]/2

or [1 – √5]/2 and [3 – √5]/2

Additional tests:

Firstly, as is the case with the golden ratio, the number y obeys the rule y^2 = y + 1

So that, the square of the number is equal to the number itself increased by 1.

And x = y + 1

So y² = x

This equality can be added to the list of other interesting equalities we started out with, namely x + y = xy = x² – y²

Now x + y = xy

y + y + 1 = xy

Inserting x = y²,

2y + 1 = y² * y

y³ = 2y + 1

Thus, the cube of the golden ratio is twice the number itself increased by 1.

This can also be derived by:

y² = y + 1

multiplying both sides by y,

y³ = y² + y --- (4)

replacing the value of y² from before,

y³ = y + 1 + y

so y³ = 2y + 1

We could similarly extend (4) to higher powers

y⁴ = y³ + y²

y⁴ = 2y + 1 + y + 1

y⁴ = 3y + 2

y⁵ = y⁴ + y³

y⁵ = 3y + 2 + 2y + 1

y⁵ = 5y + 3

y⁶ = y⁵ + y⁴
y⁶ = 5y + 3 + 3y + 2
y⁶ = 8y + 5 

and so on…

Upon closer inspection, we see that the coefficient of y and the constant term, in each successive power of y, follows a Fibonacci sequence, in which the constant term is always one term behind the coefficient of y.
Reasoning: the new coefficient of y is always the sum of the previous two coefficients of y, which follows the rule of a Fibonacci sequence, in which each term after the second term is the sum of the two terms before it. The reasoning is similar for the constant term. The Fibonacci sequence for the coefficient of y starts with the first two terms 1 and 2, while the series for the constant starts with the first two terms as 1 and 1. So:
Sequence 1: 1, 2, 3, 5, 8, 13...
Sequence 2: 1, 1, 2, 3, 5, 8, 13...