A case of Fermat's Last Theorem

[stripes]

Homework Statement

Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers.

Homework Equations

None

The Attempt at a Solution

A brute for method (i.e., considering all cases), is far too tedious. Here is the answer I gave:

It is sufficient to consider the general case 'There are no three positive integers a, b, and c such that a^3 + b^3 = c^3'. From Fermat's Last Theorem, we know that there are no integers x, y, and z such that x^n + y^n = z^n for all natural numbers n>2. Therefore, if we consider the case n=3, we have that, by Fermat's Last Theorem, there are no positive integers that satisfy the original general case a^3 + b^3 = c^3. Therefore, we cannot find two perfect cubes of positive integers whose sum is a positive perfect cube less than 1000.

Other than that, I know there Euler's proof of this statement, but it is too involved for my level. It also seems that the answer I gave is too simple. This course is a computer science course where we study theoretical foundations of comp. sci., and I have taken a real analysis course in the past. In that real analysis course, the proof above would have probably sufficed, but it seems that, from what we've learned in the comp. sci. course, this isn't the right way to approach the problem.

The assignment is due now and I will be submitting it, but I wanted to know if my "proof" is appropriate here. Thanks in advance!

 

[Dick]

If this is computer science course, then you have a computer, yes? This isn't framed like a deep question. You are just asked about numbers less that 1000. The brute force solution is easily done. If you write a program. There really aren't that many cubes less than 1000.

 

[stripes]

Never thought about that. I have to write the program in Maple, and since I am an absolute beginner at Maple programming, I wasn't able to complete my exhaustive program in time. I hope my "proof" is good enough. Thanks for your input.