From: jr@redmink.demon.co.uk (John R Ramsden)
Subject: Re: You're WRONG!
Date: Wed, 21 Mar 2001 17:53:49 GMT
Newsgroups: sci.math
Summary: No four squares in arithmetic progression 

Gerry Myerson <gerry@mpce.mq.edu.au> wrote:
>
> ALL YOUR BASE ARE BELONG TO US <spamme@lots> wrote:
> >
> > Of COURSE four integer squares can be in arithmetic progression!
> > 
> > Consider 0, 0, 0, and 0!
>
> I imagine you know what I meant. 
>
> Are there any helpful answers? 

I can't find Gerry's original post, but in this he asked if there
wasn't a simpler proof than the one in L J Mordell's Diophantine
Equations that four squares of integers can't be in arithmetic
progression (apart from trivial solutions such as all zeros).

Well the following is fairly short and, unlike the one in Mordell,
self-contained. As usual I use the Lucas Lemma, which asserts that
if integers X, Y, Z, T satisfy X.Y = Z.T then there must be a set
of integers p, q, r, s, t with (p, s) = (q, r) = 1 such that
X, Y, Z, T = p.q.t, r.s.t, p.r.t, q.s.t. In particular, if any
or all of X, Y, Z, T are known to be relatively prime then we can
assume t = 1.

As in Mordell, we represent the conditions of the problem by the
following pair of equations in which x, y, z, t are integers:

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

WLG we can assume that x, y, z, t are coprime, and considering the
equations mod 8 shows that in this case x, y, z, t are all odd.

Adding the equations gives:

   x^2 - z^2  =  y^2 - w^2

from  which there must be integers p, q, r, s with (p, s) = (q, r) = 1
and (WLG) p.q.r odd, s even such that:

   x, z, y, w  =  p.q + r.s,  p.q - r.s,  p.r + q.s,  p.r - q.s

If s = 0 then p = q = r = 1. Otherwise, subtracting the equations
gives:

   x^2 - w^2  =  3.(y^2 - z^2)

which upon replacing the preceding expressions for x, z, y, w gives:

   (q^2 - r^2).p^2 - 2.p.q.r.s - (q^2 - r^2).s^2  =  0

Treating this as a quadratic in p/s, the N&S condition for rational
p/s is that (q.r)^2 + (q^2 - r^2)^2 is the square of an integer.
So, since q.r is odd, this means there must be integers m, n with
(m, n) = 1 and m + n odd such that:

   q.r,  q^2 - r^2  =  m^2 - n^2,  m.n

The first of these implies that there must be integers X, Y, Z, W
with (W, X) = (Y, Z) = 1 such that:

   q, r, m + n, m - n  =  X.Y, W.Z, X.Z, W.Y

and replacing these in the second gives:

   (X^2 + 2.W^2).Y^2  =  (2.X^2 + W^2).Z^2

Since (Y, Z) = 1, the latter implies that for some integer T:

   X^2 + 2.W^2,  2.X^2 + W^2  =  T.Z^2,  T.Y^2   [*]

which is equivalent to the following, in which (X, W) = 1
implies that |T| must divide 3:

   3.X^2,  3.W^2  =  T.(-Z^2 + 2.Y^2),  T.(2.Z^2 - Y^2)

In view of [*] we must have T > 0 and then, by consideration
mod 8, the only possibility is T = 3 (since W.X.Y.Z is odd).

But then [*] is of the same form of the original pair, with
smaller values of the variables (left as an exercise) i.e.:

   X^2 + Z^2,  Y^2 + W^2  =  2.Y^2,  2.Z^2


Cheers

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John R Ramsden    (jr@redmink.demon.co.uk)
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The new is in the old concealed, the old is in the new revealed.
   St Augustine.
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