No Four Squares In Arithmetic Progression

Dickson's "History of the Theory of Numbers" presents a proof that 
there do not exist four squares in arithmetic progression, but I'm 
not very happy with it.  I've also received email from various people 
saying they didn't understand the proof.  Everyone seems to agree 
that the difficulty with the proof (which Dickson attributes to 
Bronwin and Furnass) is at this point:

  ... from the relations

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

  we know there are integers a,b,c,d such that
 
                     y+x = 2ab     y-x = 2cd     
                     z+y = 2ac     z-y = 2bd
                     w+z = 2bc     w-z = 2ad  

I really don't see why there have to be four integers satisfying all 
six of these relations.  The rest of the proof relies entirely on 
this step.  I'd be interested if anyone could explain why all six of 
these relations must be satisfied by four integers a,b,c,d.  Maybe 
someone with access to a reference library could check Dickson's 
sources to be sure he got it right.

Anyway, after giving up on this approach, I decided to try proving the
proposition on my own.  Here's what I came up with:

Suppose there exist four squares A^2, B^2, C^2, D^2 in increasing 
arithmetic progression, i.e., we have B^2-A^2 = C^2-B^2 = D^2-C^2.  
We can assume the squares are mutually co-prime, and the parity of 
the equation shows that each square must be odd. Then we have co-prime
integers u,v such that A=u+v, C=u-v, u^2+v^2=B^2, and the common 
difference of the progression is (C^2-A^2)/2=2uv.
 
We also have D^2 - B^2 = 4uv, which factors as [(D+B)/2][(D-B)/2] = uv.
The two factors on the left are co-prime, as are u and v, so there 
exist four mutually co-prime integers a,b,c,d (exactly one even) such 
that u=ab, v=cd, D+B=2ac, and D-B=2bd.  This implies B=ac-bd, so we 
can substitute into the equation u^2 + v^2 = B^2  to give 
(ab)^2 + (cd)^2 = (ac-bd)^2.
 
This quadratic is symmetrical in the four variables, so we can assume
c is even and a,b,d are odd.  From this quadratic equation we find that 
c is a rational function of SQRT(a^4 - a^2 d^2 + d^4), which implies 
there is an odd integer m such that  a^4 - a^2 d^2 + d^4  =  m^2.
 
Since a and d are odd there exist co-prime integers x and y such that
a^2 = k(x+y) and d^2 = k(x-y), where k=+-1.  Substituting into the 
above quartic gives x^2 + 3y^2 = m^2, from which it's clear that y must 
be even and x odd.  Changing the sign of x if necessary to make m+x 
divisible by 3, we have 3(y/2)^2 = [(m+x)/2][(m-x)/2], which implies 
that (m+x)/2 is three times a square, and (m-x)/2 is a square.  Thus 
we have co-prime integers r and s (one odd and one even) such that 
(m+x)/2=3r^2, (m-x)/2=s^2, m=3r^2+s^2, x=3r^2-s^2, and y=+-2rs.

Substituting for x and y back into the expressions for a^2 and d^2 (and
transposing if necessary) gives a^2=k(s+r)(s-3r) and d^2=k(s-r)(s+3r).
Since the right hand factors are co-prime, it follows that the four
quantities (s-3r), (s-r), (s+r), (s+3r) must each have square absolute 
values, with a common difference of 2r.  These quantities must all have 
the same sign, because otherwise the sum of two odd squares would equal 
the difference of two odd squares, i.e., 1+1=1-1 (mod 4), which is 
false.

Therefore, we must have |3r| < s, so from m=3r^2+s^2 we have 12r^2 < m.
Also the quartic equation implies m < a^2+d^2, so we have the inequality 
|2r| < |SQRT(2/3)max(a,d)|.  Thus we have four squares in arithmetic
progression with the common difference |2r| < |2abcd|, the latter being 
the common difference of the original four squares.  This contradicts 
the fact that there must be a smallest absolute common difference for 
four squares in arithmetic progression.

This theorem can be used to answer other questions as well.  For
example, H. Jurjus asked whether there are rational numbers p, q such
that (p^2, q^2) is a point on the hyperbola given by (2-x)(2-y) = 1 
with  (p^2, q^2) not equal to (1,1).  The answer is no.  To see why,
suppose p=a/b and q=c/d (both fractions reduced to least terms).  
Then if (p^2,q^2) is on the hyperbola we have

                 2b^2 - a^2      2d^2 - c^2
                 ----------      ----------   =   1
                     b^2             d^2

Since b^2 is coprime to 2b^2-a^2 and c^2 is coprime to 2d^2-c^2 it 
follows that  b^2 = 2d^2 - c^2  and  d^2 = 2b^2 - a^2.  Rearranging 
terms give

         b^2 - d^2 = d^2 - c^2         d^2 - b^2 = b^2 - a^2

Together these equations imply that a^2, b^2, d^2 and c^2 are in 
airthmetic progression, which is impossible.