Talk:Zero and One are the only Consecutive Perfect Squares

But 0 and 1 are both perfect squares. --Matt Westwood 19:25, 4 February 2009 (UTC)

Also, just because two numbers are distinct doesn't mean that their product is not a perfect square. $4\cdot 16=64=8^2$. And Matt, might we move the page to Zero and One are the only Consecutive Perfect Squares? --Cynic (talk) 22:10, 4 February 2009 (UTC)

By all means. Unless you want to include Gaussian integers, in which case $-1$ and $0$ are also consecutive perfect squares. Hm. Haven't thought as to whether it applies in the complex plane. Might need to be thought about.

Whatever, I'd like to leave Zero and One are the only Consecutive Perfect Squares to the originator of this page. Fair play, by the way. --Matt Westwood 22:48, 4 February 2009 (UTC)

Well, in the process of trying to get this to be valid, I considered the complex case, and determined that since the complex numbers under <= are not a totally ordered set, saying two numbers which are complex (but not real) are consecutive doesn't make sense. As such, I think the case can be ignored. (Yes, you could say they are consecutive if either their real part or their imaginary part are consecutive, but it doesn't really make sense.) --Cynic (talk) 02:00, 14 March 2009 (UTC)

New proof
Very nice, Lord_Farin. --Dfeuer (talk) 03:01, 27 February 2013 (UTC)


 * Thanks. It just came to me when I saw the page title in Recent Changes. &mdash; Lord_Farin (talk) 07:46, 27 February 2013 (UTC)