Content-Type: text/shitpost


Subject: A little algebraic thingy
Path: you​!your-host​!wintermute​!wikipedia​!hardees​!m5​!plovergw​!shitpost​!mjd
Date: 2018-01-23T19:53:46
Newsgroup: rec.pets.math.diophantine-equation
Message-ID: <297a95dc76b6d052@shitpost.plover.com>
Content-Type: text/shitpost

A few days ago I was wondering if there are any (nontrivial) integer solutions of $$a^2 - ab + b^2 = 1\tag{$\spadesuit$}$$

but I couldn't do it in my head. I had the idea it would be pretty easy if I tried on paper, and yes, it was one of those ones where you don't even really have to think, you just push the symbols around.

From !!(\spadesuit)!!, adding or subtracting !!ab!!, we get both $$\begin{align} a^2 + b^2 & = 1 + ab \\ (a-b)^2 & = 1 - ab \end{align}$$

and since in both cases the left sides are non-negative, we have both !!1+ab\ge 0!! and !!1-ab \ge 0!!, so !!-1\le ab \le 1!!, and we are done.

I thought about it a little more and decided that perhaps a more elegant way to put it was: Multiplying !!(\spadesuit)!! by 2, we get $$\begin{align} 2a^2 - 2ab + 2b^2 &= 2 \\ a^2 + b^2 + (a-b)^2 &= 2 \\ \end{align}$$

and then since we have three squares that sum to 2, one must be zero and the rest must be 1.

Probably there is a nice geometric proof also.