From: Timothy Murphy <tim@maths.tcd.ie>
Newsgroups: sci.math.research
Subject: Re: Diophantine eq.
Date: 22 Mar 1996 18:33:32 -0000

PROBLEM_WITH_INEWS_DOMAIN_FILE!jbuddenh@uunet.uu.net (Jim Buddenhagen) writes:

>Hauke Reddmann (fc3a501@rzaixsrv1.uni-hamburg.de) wrote:
>: Can the two equations
>: a^2+b^2=c^2
>: a^2+(2*b)^2=d^2
>: simultaneously be fulfilled
>: with _nonzero_ integers?

>For rational solutions, suppose b is non-zero, the problem is then

>x^2+1 = square   and   x^2+4 = square

>Put x=(r-1/r)/2 and get the single equation

>r^4 + 14*r^2 + 1 = square, an elliptic curve of rank 0 over Q (per APECS) 
>and (finite) torsion only with r=+-1 and r=0.  This gives no non-zero
>solutions for the original problem.

There is a more simple-minded argument by infinite descent, 
which is I think equivalent.
(Apologies if someone already said this.)

We can suppose gcd(a,b) = 1, and that a,b,c,d > 0.
By considering residues modulo 4,
it is clear that a, c and d are odd and b even.

By the standard (Pythagorean?) solution to

a^2 + b^2 = c^2

we can write

a = u^2 - v^2, b = 2uv, c = u^2 + v^2,

with gcd(u,v) = 1.

Similarly, from

c^2 + b^2 = d^2

it follows that

c = U^2 - V^2, b = 2UV, d = U^2 + V^2,

with gcd(U,V) = 1.

Thus

u^2 + v^2 = U^2 - V^2, uv = UV.

It follows from the second equation that we can write

u = xz, v = yt, U = xt, V = yz,

where x, y, z, t are mutually co-prime.
The first equation now reads

y^2(t^2 + z^2) = x^2(t^2 - z^2).

But now

gcd(x,y) = 1, gcd(t^2 + z^2, t^2 - z^2) = 1 or 2.

Thus either

x^2 = t^2 + z^2, y^2 = t^2 - z^2,

or

2x^2 = t^2 + z^2, 2y^2 = t^2 - z^2.

In the first case we have

y^2 + z^2 = t^2, y^2 + 2z^2 = x^2,

while in the second case we have

z^2 + 2y^2 = t^2, z^2 + y^2 = x^2.

Thus in either case we have another solution of the original two equations.

Finally, it is easy to see that in the first case

z < t <= v < b,

while in the second case

y < t <= v < b.

Thus in either case we have a 'smaller' solution,
as measured by the size of b.
This leads to a contradiction, by infinite descent.




-- 
Timothy Murphy  
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland