No Four Rectangles In Line?

The note entitled Discordance Impedes Square Magic included a
table of edge lengths for rectangles with rational edge lengths but
not necessarily rational diagonal, and shows that apparently no four
such rectangles exist with areas in arithmetic progression.  This 
calls to mind the fact that there are no four squares with areas in 
arithmetic progression.  So, I'm going to boldly make a completely 
unsupported generalization of these two facts.

First, notice that any rectangle with integer edge lengths a,b 
corresponds uniquely to a point on a plot of (a^2 - b^2) versus 
(a^2 + b^2), as shown below.

            |
     a^2-b^2|        /
            |      /
            |    /
            |  /
  __________|/_____________
            |\       a^2 + b^2
            |  \
            |    \
            |      \
            |        \

The right-hand region in between the two diagonal lines represents 
the rectangles with a >= b.  Lines of constant area are hyperbolas,
asymptotic to the diagonal lines.

The squares are the rectangles with a=b, so those all lie along
the horizontal axis.  On the other hand, a set of rectangles 
with a common diagonal all lie along a single vertical line.

So here's my reckless conjecture:  There do not exist four points 
on any single line (of any slope) in this plane corresponding to 
rectangles with areas in arithmetic progression.  Thus, the claim 
is that there do not exist constants A,B and non-zero integers aj, 
bj for j=1,2,3,4 (assume aj >= bj) such that

    (aj^2 - bj^2) =  A(aj^2 + bj^2)  +  B     j=1,2,3,4        (1)

and such that the four products (aj bj), j=1,2,3,4 are in arithmetic 
progression.  Obviously in the special case A=B=0 this requires 
aj=bj, and so the claim in this case is that the four squares aj^2 
can't be in arithmetic progression, which is known to be true.  This 
corresponds to four point on the horizontal line of the a^2 + b^2 
axis.

It's less convenient working in these coordinates for vertical lines,
because then the coefficient A would be infinite.  A more natural set 
of coordinates for this problem would be the diagonal lines shown on 
the above figure, because they represent the degenerate cases when 
b=0 or a=0, respectively, which are not proper rectangles.

Using those axes as our coordinates, we see that they represent the 
a^2 and b^2 projections, and straight lines in these coordinates map 
to straight lines in the original coordinates.  Specifically, if we 
denote a straight line in the a^2, b^2  coordinates as

                  aj^2 = M bj^2  +  K                       (2)

then this corresponds to a straight line in the form of (1) with the
coefficients
                  M - 1                2K
            A  =  -----         B  =  -----
                  M + 1               M + 1

The advantage of the MK coordinates is that they are not singular 
for any proper aspect ratio.  So, the original claim can be expressed 
by saying there are no constants M,K and positive integers aj,bj, 
j=1,2,3,4 (with aj >= bj) such that that (2) is satisfied for all
j and such that the four products (aj bj), j=1,2,3,4 are in 
arithmetic progression.

It might be possible to prove (or disprove) some special cases of
this conjecture.  For example, consider the horizontal lines in 
the original drawing.  These have the form a^2 = b^2 + K for some 
constant K.  We know that with K=0 this is just the well-known 
theorem on four squares in arithmetic progression.  With K not 
equal to zero we have other sets of rectangles, such as the line 
K=105, which contains the lattice points (a,b) = (53,52), (19,16), 
(13,8), (11,4).  Each of these points corresponds to a factorization 
of 105.  Obviously the areas of these particular rectangles are not 
in arithmetic progression.  Can we prove that four such areas can 
never be in arithmetic progression for any given K?

In summary, there exist sets of three rectangles with integer edges 
lengths and the same diagonal (not necessarily rational) such that 
their areas are in arithmetic progression, but it appears that no 
FOUR such rectangles exist.  This is interesting, because it leads 
to a generalization of the well-known fact that no four squares have 
areas in arithmetic progression.  The general conjecture is that 
there do not exist constants A,B and integers xi,yi (i=1,2,3,4) 
such that 
                yi^2 = Axi^2 + B     i=1,2,3,4

with the four products (xi yi) in arithmetic progression.
Obviously with A=1,B=0 this is the well-known theorem on four 
squares.  On the other hand, with A=-1 and B equal to an arbitrary 
constant (K), this gives the (seeming) fact that no four rectangles 
(with integer edges) on a common diagonal can have areas in 
arithmetic progression.  In other cases, such as setting A=D,B=1 
we see that the x,y pairs are solutions of a single Pell equation.