Very 'pell' liang de problem
by shiningsunnyday, Sep 28, 2016, 3:34 PM
Interm NT wrote:
A consecutive Pythagorean triple is a Pythagorean triple of the form
,
and
positive integers. Given that
,
, and
form the third consecutive Pythagorean triple, find
.







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 











Case 1:


This generates




Case 2:


This generates




Therefore, the second such triple will have




Remark
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

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


