Prove that for any three positive integers a, b, c such that the product of any two is one less than a square,
arithmetic/square.products New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Prove that for any three positive integers a, b, c such that the product
of any two is one less than a square,
ab+1 = x^2
ac+1 = y^2
bc+1 = z^2
there is always a fourth positive integer d whose product with each of
the first three is also one less than a square:
ad+1 = w^2
bd+1 = u^2
cd+1 = v^2
Show solution
[ Editor: posed by Phil Gibbs <gibbs@eurocontrol.fr>, who offered a
crate of haddock to the first solver. ]
Keep the haddock. I was just trying to figure out one case,
and ran into a problem finding the right d. Put a=6, b=8, c=28.
ab+1 = 6*8 +1 = 7^2
ac+1 = 6*28+1 = 13^2
bc+1 = 8*28+1 = 15^2
So in this particular case one wants to make each of
6d+1, 8d+1, 28d+1
a square. Now 6d+1 is square exactly for d = 2p*(3p+-1),
and 8d+1 is square exactly for d = q*(2q+-1), and finally
28d+1 is square exactly for d = r*(7r+-1).
[easy to show, and easy to verify experimentally...]
-------------------------
So to find d one has "only" to find p,q,r such that the
following "triple equation" holds:
(*) 2p*(3p+-1) = q*(2q+-1) = r*(7r+-1) .
[with one of the 8 possible choices of +- sign]
Even equating two of these amounts to a quadratic diophantine
equation, of the type usually approached via "pellian" methods.
That would be doable, but with all three needing to be equal, it
looks hard.
Maybe someone could find an appropriate d via computer search...
Or maybe someone already knows how to solve equation (*) in integers
p,q,r by some other method....
-- Dan Cass <dmc@sjfc.edu>
The triple equation reduces to the pair of simultaneous Pellians
4x^2 - 3y^2 = 1
7x^2 - 3z^2 = 4
where
x = 6p +- 1, y = 4q +- 1, z = 14r +- 1.
One can use Baker's method to show that any solution has log(x) < 63.1
A computer search for small values of x suggests that the only solutions
are x = 1, y = 1, z = 1. One can now use a 'sieving' technique to prove
that these are the only solutions with x < e^69.66 and hence the only
solutions. The computations took a few seconds. The method is
described in detail in two papers of mine:
R.G.E. Pinch, Simultaneous Pellian equations,
Math Proc Cambridge Philos Soc 103 (1988) 35-46
R.G.E. Pinch, Squares in quadratic progression,
Math Comp 60, 202 (April 1993) 841-845
-- Richard Pinch <rgep@newton.cam.ac.uk>
[ Editor: the second Pellian does not follow from Cass's equations.
With Pinch's x, y, z we have x^2 = 6d+1, y^2 = 8d+1 and z^2 = 28d+1.
The first equation checks: 4(6d+1) - 3(8d+1) = 1. But eliminating d
between x^2 and z^2 gives 14x^2 - 3z^2 = 14 - 3 = 11, not
7x^2 - 3z^2 = 4. And the triple 6, 8, 28 does have a fourth member:
d = 5460, for which 6d+1 = 32761 = 181^2, 8d+1 = 43681 = 209^2 and
28d+1 = 152881 = 391^2 (x = 181, y = 209, z = 391 satisfy
4x^2 - 3y^2 = 1 and 14x^2 - 3z^2 = 11).
This d comes from a formula that answers Gibbs's question for every
triple. Writing ab+1 = x^2, ac+1 = y^2, bc+1 = z^2 as in the problem,
put d = a+b+c+2abc+2xyz, which is a positive integer. Expanding and
substituting x^2, y^2, z^2 shows
ad+1 = (az+xy)^2, bd+1 = (by+xz)^2, cd+1 = (cx+yz)^2.
For example (az+xy)^2 = a^2(bc+1) + 2axyz + (ab+1)(ac+1)
= 2a^2bc + a^2 + ab + ac + 2axyz + 1 = a(a+b+c+2abc+2xyz) + 1.
For 6, 8, 28: d = 42 + 2688 + 2730 = 5460, and az+xy = 90+91 = 181,
by+xz = 104+105 = 209, cx+yz = 196+195 = 391. ]