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EMAT 6600

Problem Solving

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An algebraic Exploration

Problem: Prove that x² – y² = A³ always has integer solutions (x,y) whenever A is a positive integer.

Solution

x² – y² = A³

(x + y)(x – y) = A³ = A² * A

now if we consider either of the following,

x + y = A² and x – y = A

we then have

x = A² – y and x = A + y

A² – y = A + y

A(A – 1) = 2y

hence, y = A(A –1)/2.

or

x + y = A and x – y = A²

similarly, we have

x = A – y and x = A² + y

A – y = A² + y

A – A² = 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.¨

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