Diophantine ntuples and Their Duals 

Diophantus considered rational solutions of the system of equations 

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An infinite class of solutions (though not all solutions) can be developed from formulas for the areas of triangles expressed in terms of generalized “edge lengths”. Given an arbitrary triangle, let (u_{1},v_{1}), (u_{2},v_{2}), and (u_{3},v_{3}) denote the components of the three edge vectors going clockwise around the triangle, in terms of some Cartesian coordinate system. Since these vectors form a closed loop, we have u_{1}+u_{2}+u_{3} = 0 and v_{1}+v_{2}+v_{3} = 0. Now let s_{j} denote the generalized “edge length” defined as 

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for given constants a,b,c. (If we take a = c = 1 and b = 0, these are simply the actual lengths of the edges.) The area A of the triangle is given by any of the three equivalent expressions 

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By the generalized version of Heron’s formula, the area of the triangle is also given by 

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If we define s = (s_{1} + s_{2} + s_{3})/2, this can be expressed more compactly as 

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To show how this formula gives Diophantine triples, let us define S_{j} = s_{j}^{2}, and then expand the products on the right hand side and multiply through by 4 to give 

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This can be rewritten in several equivalent forms, including 

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These three expressions show that the set of numbers {S_{1},S_{2},S_{3}} constitutes a solution of the problem of Diophantus mentioned above, with k = (b^{2}ac)(2A)^{2}. Incidentally, the above relations can also be written in the explicitly symmetrical form 

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In the special case when k = 1 the integers {S_{1},S_{2},S_{3}} are called a Diophantine triple, and equation (1) reduces to 

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To illustrate, consider the triangle depicted below 


The area of this triangle is A = 1/2, and the edge vectors are (u_{1},v_{1}) = (0,1), (u_{2},v_{2}) = (1,1), and (u_{3},v_{3}) = (1,0), so the S_{j} values are 

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If we now choose a,b,c such that the discriminant of the quadratic form defining the S_{j} values is unity, then these values will constitute a Diophantine triple. Therefore we need only choose integers a,b,c such that b^{2} – ac = 1, which is equivalent to ac + 1 = b^{2}. Of course this condition is independent of the sign of b, so given any Diophantine 2tuple {a,c} with ac + 1 = b^{2}, we have the Diophantine 3tuples {a, c, a+c2b} and {a, c, a+c+2b}. This is easily verified by noting that 

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The same result can also be reached directly by simply solving the basic identity (2) for any one of the S_{j} values in terms of the other two. For example, if we solve (2) for S_{2} we get 

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This shows that if S_{1},S_{2},S_{3} satisfy (2) then S_{1}S_{3} + 1 must be a square, and by symmetry so must be S_{1}S_{2} + 1 and S_{2}S_{3} + 1, which confirms that every solution of (2) in integers represents a Diophantine 3tuple. These are called regular Diophantine 3tuples, to distinguish them from Diophantine 3tuples that are not solutions of (2). The above discussion also shows that every regular Diophantine 3tuple can be found by starting with a 2tuple {a,c} and then extending it to a 3tuple as described above. 

It’s interesting that regular Diophantine 3tuples can be represented by triangles in two very different ways. First, as we’ve just described, they can all be represented by the generalized Heron formula based on the same primitive triangle depicted above, by simply choosing different coefficients a,b,c for the quadratic form of the generalized edge length definition such that b^{2} – ac = 1. But they can also be represented for any fixed choice of coefficients a,b,c by choosing different triangles of area 1/2 with rational edge vectors. To illustrate, consider the simplest nontrivial Diophantine triple {1,3,8}. In order to represent this in terms of the primitive triangle shown above, we would select coefficients of the generalized length function a = 1 and c = 3, and then we must have b = ±2. Thus our generalized squared edge length function is 

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Inserting the edge vector components for our standard primitive triangle (depicted above), we get S_{1} = 3, S_{3} = 1, and S_{2} = 0 or 8. But there is another way of representing the triple {1,3,8}, using a standard edge length definition and selecting a suitable triangle. Let us take b = 1/2 and a = c = 0 as our standard coefficients, so the square edge lengths are simply 

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Now we need a triangle whose area is A = 1, and for this we can take the triangle depicted below. 


With these edge vector components and generalized length definition we have S_{1} = 1, S_{2} = 3, and S_{3} = 8, and it’s easy to verify from equation (0) that the area is 1. In this same way we can represent any regular Diophantine triple by a triangle of area 1 with integer edge vector components, always using the standard generalized length definition given by equation (3). 

The fact that regular Diophantine triples can be represented by a standard triangle with variable quadratic forms, or alternatively by a standard quadratic form with variable triangles, is reminiscent of how quantum mechanical systems can be represented by a constant state vector with variable measurement operators, or alternatively by standard measurement operators with variable state vectors. 

So far we’ve focused mainly on the case when (b^{2}ac)(2A)^{2} = 1, corresponding to Diophantine triples. If, instead, we choose a quadratic form such that 4(b^{2}ac)A^{2} = 1, then S_{1}, S_{2}, S_{3} constitute what might be called an antiDiophantine triple, because each of their pairwise products reduced by unity is a square. For this purpose we replace (3) with the definition 

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Just as the s_{j} based on (3) are the geometric mean of u_{j} and v_{j}, the definition of s_{j} based on (4) is the rootmeansquare. In this case we have a = c = 1/2 and b = 0. We again want a triangle with squared area equal to 1, and with integer edge components (with the same parity), so we can use the same triangle we used for the previous case. Inserting the edge vector components u_{1} = v_{1} = 1, u_{2} = 1, v_{2} = 3, u_{3} = 2, v_{3} = 4, into equation (4), we get the antiDiophantine triple S_{1} = 1, S_{2} = 5, S_{3} = 10. This is the dual of the Diophantine triple {1,3,8}. A summary of these dual triples is shown below. 

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One might wonder if, for any given regular triple and any given quadratic form, we can infer the components of the edge vectors of a corresponding triangle. Let us take the quadratic form (3) as an example. We know that u_{1}+u_{2}+u_{3} = 0, and similarly v_{1}+v_{2}+v_{3} = 0, so we can substitute –(u_{1}+u_{2}) and –(v_{1}+v_{2}) for u_{3} and v_{3} respectively into the last condition to give the system of equations in four variables 

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Solving the first two equations for v_{1} and v_{2}, substituting into the third, and rearranging terms, we get 

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We can then solve this quadratic for (u_{1}/u_{2}) to give 

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We recognize the argument of the square root as our previous expression for the squared area of the triangle, which in this case is 1. Inserting the values of S_{j} for this example, we get u_{1}/u_{2} = 1 or 1/3, and so this implies the u values 1,1,2 or else 1,3,4, and whichever set we choose, the v_{j} values are the other set. Using the first choice, we have u_{1} = v_{1} = 1, u_{2} = 1, v_{2} = 3, u_{3} = 2, v_{3} = 4. 

For a less trivial example, consider the Diophantine triple S = {8,21,55}. Inserting these values into equation (5) we get u_{1}/u_{2} = 2/3 or 4/7, so we take u = {2,3,5} and v = {4,7,11} to give the antiDiophantine triple S = {10,29,73}. For these triples we have the results 

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Incidentally, we could describe the usual Diophantine triples as “hyperbolic”, and the antiDiophantine triples as “elliptical”, because they correspond respectively to the cases where the discriminant b^{2} – ac of the quadratic form is positive or negative, and hence the locus described by the quadratic form is respectively a hyperbola or an ellipse. If we were interested only in rational solutions, then for any quadratic form with a rational square determinant, i.e., with b^{2} – ac = d^{2} or else ac – b^{2} = d^{2} where d is a positive rational number, we could divide through each of the basic triple equations and write them in the form 

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Thus every triangle (of any area) with rational edge vector components, and with any rational quadratic form (hyperbolic or elliptical), corresponds to a rational Diophantine or antiDiophantine triple. 

It’s natural to try to extend these triples by progressing from the generalized Heron formula for the area of a triangle to the generalized Brahmagupta formula for the area of a quadrilateral (whose vertices lie on a locus with the given quadratic form) in terms of the components of the edges. Thus we’re led to consider a 4th edge vector with components u_{4},v_{4}, and to define s_{4} and S_{4} in terms of the chosen quadratic form of these components. Of course, we now have the loop conditions u_{1 }+ u_{2 }+ u_{3} + u_{4} = 0 and v_{1} + v_{2} + v_{3} + v_{4} = 0, so in general no three of the vectors form a closed loop, and hence no subset of the S_{j} values need be a regular triple, even though they constitute a triple. 

According to the generalized Brahmagupta formula, the squared area of a quadrilateral whose vertices lie on a locus of the quadratic form a(xx_{0})^{2} + 2b(xx_{0})(yy_{0}) + c(yy_{0})^{2} = r^{2} is given by 

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which of course reduces to Heron’s formula for the area of a triangle if any of the edge vectors are null, i.e., if s_{j} = 0. The right hand side can be expanded and the resulting terms collected in several different ways. All but one of the terms involve only the squared parameters s_{j}^{2}. The exception is the term 8s_{1}s_{2}s_{3}s_{4}. The area formula for a general quadrilateral (with vertices not necessarily on a locus of the given quadratic form) includes on the right hand side the term 16u_{1}u_{2}u_{3}u_{4}cos(q)^{2} where q is half the sum of two opposite angles, so we could cancel these terms if we specialized to quadrilaterals with q = p/4. However, rather than take this approach, we will continue to focus on quadrilaterals inscribed in the given quadratic form, and isolate the exceptional term in such a way that it can be absorbed into a simple condition on the area A. To do this, we arrange the terms of the product in either of the following two ways: 

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The overall expression is explicitly symmetrical in the four parameters, as is the last term on the right side of each expression, so the sum of the first two terms must also be symmetrical, as can be confirmed by expanding them. This motivates our partitioning of the terms in this way, since we seek a symmetrical expression (analogous to (1) above) involving the four S_{j} values. We’ve developed two different expressions because we want to equate the final square terms to the quantity 16(acb^{2})A^{2}, which is positive or negative depending on whether the quadratic form is hyperbolic or elliptical. Thus for these cases we impose the condition 

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which can also be written as 

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Now, for the hyperbolic case, given any triple {S_{1},S_{2},S_{3}}, which need not be regular, we can compute the value of S_{4} that satisfies this condition, which implies that the sum of the first two terms on the previous expressions must vanish. Thus we compute the value of S_{4} such that each of the following three equivalent equations is satisfied 

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Solving this quadratic condition for S_{4} we get 

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This shows that if {S_{1},S_{2},S_{3}} is a Diophantine triple, then S_{4} is an integer, because each of the three factors under the square root sign is a square. Furthermore, the three preceding equations imply that each of the quantities S_{1}S_{4} + 1, S_{2}S_{4} + 1, and S_{3}S_{4} + 1, is a square, so {S_{1},S_{2},S_{3},S_{4}} is a Diophantine 4tuple. For example, beginning with the Diophantine triple {1,3,8}, we compute S_{4} = 0 or 120. The fact that one of the S_{4} roots is zero signifies that the original triple is regular. Taking the nontrivial root S_{4} = 120, we have 

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To find the explicit expression for these three squares in this case (originally found by Arkin, Hoggatt, and Strauss in 1979), by the initial Diophantine triple conditions we have integers a,b,g such that 

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and we can evaluate the first condition involving S_{4} as 

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Hence by symmetry we have 

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Similarly for the elliptical case we have the quadratic condition represented by each of the three equivalent expressions 

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Solving this quadratic condition for S_{4} we get 

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This shows that if {S_{1},S_{2},S_{3}} is an antiDiophantine triple (not necessarily regular), then S_{4} is an integer, because each of the three factors under the square root sign is a square. Furthermore, the three preceding equations imply that each of the quantities S_{1}S_{4} – 1, S_{2}S_{4} – 1, and S_{3}S_{4} – 1, is a square, albeit of an imaginary integer (because S_{4} is negative). We will call such a set of four numbers an antiDiophantine 4tuple. As an example, beginning with the triple {1,5,10} we compute S_{4} = 168, for which we have 

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To find the explicit expression for the (negative) squares in this case, by the initial antiDiophantine triple conditions we have integers a,b,g such that 

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and we can evaluate the first condition involving S_{4} as 

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Hence by symmetry we have 

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Since these expressions give negative squares, these 4tuples do not represent antiDiophantine 4tuples in real integers, although they do in Gaussian integers. It appears that there are no antiDiophantine 4tuples in real integers. 
