Clouds, Shy Squares, and Diophantus

Do there exist 5 positive integers such that the product of any two 
is one less than a square?  For a discussion of this and related 
questions, see If ab+1, ac+1, bc+1 are squares,... .
As an aside, there's a connection between this problem and the sets 
of points called "clouds" by L. Comtet in his book "Advanced 
Combinatorics".  If we have 3 (distinct) positive integers a,b,c 
(e.g., 1,3,8) such that each of the three pairwise products ab,ac,ad 
is a "shy square" (defined as one less than a square), then obviously 
the square (abc)^2 is expressible as a product of three shy squares, 
i.e., (abc)^2 = (A^2-1)(B^2-1)(C^2-1).  

However, if we have 4 distinct integers a,b,c,d (e.g., 1,3,8,120) 
such that each of the SIX pairwise products is a shy square, then 
the square (abcd)^2 must not only be expressible as a product of 
four shy squares, it must be so expressible in THREE distinct ways, 
corresponding to the three ways of choosing four out of the six 
pairwise products such that each of a,b,c,d appears exactly twice.
For example, with a=1,b=3,c=8,d=120 we have

    (2880)^2  =  (2^2 - 1)(3^2 - 1)(19^2 - 1)(31^2 - 1)
              =  (2^2 - 1)(5^2 - 1)(11^2 - 1)(31^2 - 1)
              =  (3^2 - 1)(5^2 - 1)(11^2 - 1)(19^2 - 1)

Now, if we imagine 5 distinct integers a,b,c,d,e such that each of 
the TEN pairwise products is a shy square, then the square (abcde)^2 
must be expressible as a product of 5 distinct shy squares in TWELVE 
distinct ways.  It isn't trivial to find numbers that are expressible 
as products of shy squares in multiple ways.  For example, no one has 
ever found ANY integer (let alone a square) that is expressible in 
the form (n^2 - 1)(m^2 - 1) in more than five distinct ways. (And 
only 6 five-way expressible numbers are known).

Anyway, the connection to Comtet's "clouds" is that if you go on to 
consider N distinct integers such that each of the N(N-1)/2 pairwise 
products is a shy square, then the square of the product of those N 
integers must be expressible as a product of N shy squares in c(N) 
distinct ways, where the first few values of c(N) are

            N    c(N)
           ---  -----
            3      1
            4      3
            5     12
            6     70
            7    465
            8   3507

This is the number of ways of choosing N out of N(N-1)/2 pairwise 
products of N numbers such that each of the individual numbers 
appears exactly twice.  Checking the Encyclopedia of Integer 
Sequences (Sloane and Plouffe) we find that this is sequence M2937, 
which Comtet defined in terms of the intersection points of N lines 
in general position in the plane.

Incidentally the question about 5 positive integers such that the 
product of any two is one less than a square goes back to Diophantus.
It's true that Diophantus only required rational numbers, rather than
restricting his domain to integers, but of course many of his results 
have relevance for integers too.  In Book III of The Arithmetica he 
treated the problem of finding three numbers such that the product of 
any two of them increased by 1 is a square, and he gave the triple 
a=x, b=x+2, c=4x+4.   Then in Problem 20, Book IV, he treated the 
problem of finding FOUR numbers such that all six pairwise products 
are 1 less than a square.  He built on his triple solution a,b,c, but 
oddly enough he *didn't* use the case x=1 (as Fermat later did to 
find 1,3,8,120).  Instead he decided to make the product of the first 
and fourth numbers equal to one less than (3x+1)^2.  I suppose he 
was just following the pattern 

  ab+1 = (1x+1)^2
  ac+1 = (2x+1)^2       bc+1 = (2x+3)^2
  ad+1 = (3x+1)^2       bd+1 =   ?          cd+1 = (6x+5)^2

Notice that setting ad+1 = (3x+1)^2 forces d to be 9x+6, which
automatically gives cd+1 = (6x+5)^2.  Diophantus didn't comment
much on this, but it's clearly systematic.  For example, if he 
went on to a 5th number e such that ae+1 = (4x+1)^2 it would 
force e to be 16x+8, which automatically gives de+1 = (12x+7)^2, 
and so on.  

Anyway, all he needs to do now is find a (rational) value of x such 
that  bd+1 =  9x^2 + 24x + 13  is a (rational) square.  He reasons 
that if we assume bd+1 = (3x+k)^2 for some integer k then we will 
have (24-6k)x + (13-k^2) = 0, and so x can be any number of the form  
x = (k^2 - 13)/(24 - 6k).  He chooses k=-4, which gives x=1/16, so
his four numbers are 1/16,  33/16,  68/16, and  105/16.  (He might
have chosen k=-11 to give x=6/5.)

Euler gave a family of such rational quintuples, including these
two examples

     1/2   5/2   6    48    44880/128881
      1     3    8   120   777480/8288641


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