Re: Diophantine Pythagoras horror

In article <pl2k1296g656v1v02akm0f19o0gtomvtfa@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
On Thu, 16 Mar 2006 23:47:09 GMT, "Peter L. Montgomery"
<Peter-Lawrence.Montgomery@xxxxxx> wrote:

In article <dvbrs1$i03$3@xxxxxxxxxxxxxxxxxxxxxxxxxx>
Hauke Reddmann <fc3a501@xxxxxxxxxxxxxx> writes:
Let a=ix+j,b=kx+l,c=mx+n.
Find i,j,k,l,m,n (integer>0) such that the
six solutions (in x) of (nonsimultaneous!!)
a^2+b^2=c^2
b^2+c^2=a^2
c^2+a^2=b^2
are all integers.

If you pick i = k + m and j = l + n, then a = b + c (as a polynomial).
All of a^2 + b^2 - c^2, b^2 + c^2 - a^2, c^2 + a^2 - b^2
split into two linear factors (in x) over Q. It shouldn't be hard to
restrict k, l, m, n so all roots are integers.

Nice idea.

You tamed a monster.

Applying your plan, here is one class of solutions:

Let l, n be positive integers, both odd, but otherwise arbitrary.

Define m, k, i, j by:

m = 1

k = 1

i = 2

j = l + n

Then each of the equations

a^2 + b^2 - c^2 =0

b^2 + c^2 - a^2 = 0

c^2 + a^2 - b^2 = 0

has two integer solutions for x.

But there are only three distinct x values:
-l, -n, and -(l+n)/2, each satisfying two equations.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


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