EMAT 6600

Problem Solving

**An algebraic
Exploration**

**Problem**:
Prove that x^{2} – y^{2} = A^{3} always has
integer solutions (x,y) whenever A is a positive integer.

x^{2} – y^{2} = A^{3}

(x + y)(x – y) = A^{3} = A^{2} * A

now if we consider either of the following,

x + y = A^{2} and x – y = A

we then have

x = A^{2} – y and
x = A + y

A^{2} – y = A + y

A(A – 1) = 2y

hence, y = A(A –1)/2.

or

x + y = A and x – y = A^{2}

similarly, we have

x = A – y and x = A^{2} + y

A – y = A^{2} + y

A – A^{2} = 2y

A(1 – A)/2 = y

Now, equate the values of y

A(1 – A) = A(A – 1)

1 – A = A – 1

2 = 2A Þ 1 = A

thus, A is a positive integer.¨