Newsgroups: rec.puzzles
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: 1995/05/12
Subject: Re: integral pythagorean theorem puzzle...

> Aaron Koller <15821...@msu.edu> wrote:

> > a,b, and c are positive integers greater than 2.  What is the
> > GREATEST value for b (in terms of a) such that a^2+b^2=c^2?

Does this really need a spoiler warning?

We have a^2=c^2-b^2; since the difference between adjacent squares
increases with b (and the distance between nonadjacent squares
increases faster) the best we can do is c=b+1.

a^2=(b+1)^2-b^2=2b+1, so b=(a^2-1)/2.  Clearly this is an integer if
and only if a is odd.

If a is even, then, we can do no better than c=b+2.  Then
a^2=(b+2)^2-b^2=4b+4, so b=a^2/4 - 1, which is an integer.

Dan
Hoey@AIC.NRL.Navy.Mil