My interest in hyperdeterminants arose from investigating a very old problem in number theory that originated with Diophantus himself. In his books on algebra probably written around 250 AD, Diophantus looked at many equations which he tried to solve in rational numbers. One such problem was to find sets of rational numbers such the product of any two is one less than a square. I have no idea what motivated him to consider this particular puzzle. It has no obvious interpretation in geometry or any other practical application. He seems to have just invented it as an abstract quastion. However, I do know that it was well chosen because this problem has hidden symmetries of enormous interest even today and its mysteries are still not fully revealed.
Diophantus provided examples of triplets and quadruples of rational numbers that solved his problem. A neater example was provided by Fermat who read Diophantus but became more interested in integer solutions to equations than rational ones. He observed that 1,3,8,120 has the required property with the products being one less than the squares of 2,3,4,11,19 and 31. We still dont know if there are 5 non-zero integers with this property although Dujella has shown that there can be at most a finite number of them. Euler found that a fifth rational number (777480/8288641) can be added to Fermat’s quadruple giving a set of five. A few years ago I carried out a computer search and found some rational diophantine sextuples of which the smallest is this one
11/192 35/192 155/27 512/27 1235/48 180873/16
There are still plenty of related unsolved problems and a growing literature on the subject for any number theorist with time on their hands. They might start at the list of references maintained by Dujella at http://web.math.hr/~duje/ref.html
Of more interest to me these days are the hidden symmetries in this problem. These first become evident when you attack the probelm directly. To find two numbers with the property is easy. We just want a and b with
ab = x2 -1
In rationals we can just take any non-zero numbers a and x and solve b = (x2-1)/a for a complete solution over rationals. A third number would take the form c = (y2-1)/a , with the requirement that bc = z2 -1
(x2-1)(y2-1)/a2 = z2 -1
Because the largest part of each side is nearly a square it is natural to look for a perturbation solution with
z = (xy + t)/a
Eliminating z in favour of t and simplifying reduces to
x2 + y2 + t 2+ 2xyt = a2 + 1
The first thing to notice about this equation is that it is quadratic in any of the variables x,y,t. This means that if we have one solution in integers or rationals, then we can find another by looking for the other root of the quadratic for any of the three variables. This can be repeated to give an infinite nunber of diophantine triples a,b,c. The second observation is that the equation is symmetrical in the variables x,y,t. This is an accidental hidden symmetry that was not apparent originally and it has a useful consequence. It means that given any triplet solution a,b,c we can construct a fourth number d = (t2-1)/a and by the symmetry of the equation this must extend the triplet to a quadruplet with the property of Diophantus.
Although the equation shows some symmetry it does not reflect the expected symmetry of the problem between the number a,b,c,d. This can be remedied by eliminating x,y and t in favour of b,c and d,
(ab+1) + (ac+1) + (ad+1) + 2 sqrt((ab+1) (ac+1) (ad+1)) = a2 + 1
=> 4(ab+1) (ac+1) (ad+1) = (a2 – ab – ac – ad – 2)2
By expanding this, cancelling terms and dividing out a factor of a2 we get an equation symmetric in a,b,c,d
P(a,b,c,d) = a2 + b2 + c2 + d2 -2ab – 2ac – 2ad – 2bc – 2bd – 2cd – 4abcd + 4 = 0
By construction this equation should be capable of being used to extend an Diophantine triple (a,b,c) to a Diophantine quaadruple (a,b,c,d). This can be seen explicitly if we can use it to solve for d when a,b,c are given. Since the equation is quadratic in d this amounts to completing the square to rewrite it as
P(a,b,c,d) = (d – a – b – c – 2abc)2 – 4(ab+1)(ac+1)(cd+1) = 0
So if (a,b,c) is a Diophantine triple meaning that (ab+1), (ac+1) and (cd+1) are integer squares, then the equation can indeed ne solved to findand integer d. That (a,b,c,d) is then a quadruple follows from the observation that (ad+1) must be a square because
P(a,b,c,d) = (a – b – c + d)2 – 4(ad+1)(bc+1) = 0
and similar identities for the other squares.
At this point it appears that the ability to extend a triple to a quadruple is the result of the fact that this polynomial can be written in a number of different ways as the difference of a square and a product of the numbers that need to be squares. This result seems to defy explanation. If this is not remarkable enough it is more surprising that there is a similar polynomial in five variables that allows any Diophantine quadruple to be extended to a rational Diophantine quintuple. I will return to that in a later post.
So where does this polynomial come from and to what mathematical principles does it owe its properties?
A partial answer to this question is that the expression is a special case of Cayley’s hyperdeterminant for a 2x2x2 hypermatrix with components
a111 = -a
a112 = 1
a121 = 1
a122 = b
a211 = 1
a212 = c
a221 = d
a222 = -1
The additional symmetries of the hyperdeterminant reduce the number of mysterious identities that it fulfills and most of those that remain arise if the derivation of the hyperdeterminant. Yet it is not obvious that this fully explains the relationships and some mystery remains.