Jump to content

Squares and Cubes


Recommended Posts

DrKrettin: that wouldn't work, as then a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 2*something odd, which can't be a square (except in the case that n = 0, a = 2, b = 0, and it still isn't a square).

Edited by uncool
Link to comment
Share on other sites

DrKrettin: that wouldn't work, as then a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 2*something odd, which can't be a square (except in the case that n = 0, a = 2, b = 0, and it still isn't a square).

 

Yes - I'd decided that approach was getting nowhere. There doesn't seem to be an analytical way of doing this.

Link to comment
Share on other sites

Nice. Do you know whether that is a unique answer, or are there other larger ones?

 

No idea - ran through a few numbers in my head on my ride to work today. I cannot think of an analytical method - or any way other than sieve tbh

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.