The first such triple is obviously Ack. Bounding + mods + guess/check approaches are dead ends and bashy. We want to relate to since having as the RHS tells us nothing about the possibilities, so let's make the substitution

simplifies to so we desire to be a perfect square. Note that the case gives us the first triplet. Since are relatively prime, it must be the case both are odd perfect squares, with one of them having an extra factor of

Case 1: where are odd

This generates our well-known Pell's Equation, with initial solution and subsequent solution of which seems so large... Let's keep looking.

Case 2: where are odd

This generates with an initial solution of so candidates for solutions are which yield no solutions, and which works!

Therefore, the second such triple will have and the third triple will be and solving gives us

The official solution just manipulates the equation into and uses Pell's directly, LOL. Sorry for being a scrub and making the sol unnecessarily long. :\

First time solving a problem with Pell's though - really good stuff

I should probably work through a PEN or PEN-equivalent resource over the winter or something, both for USAMO (hopefully!) and in case I actually get accepted into a non-empty subset of the camps I'm applying to this year: so I can learn some really awesome stuff like elliptical curves over the summer.