Arranging the Solutions of f(x+y+z) = xyz

For any quadratic polynomial f(s) = as² + bs + c with integer coefficients a,b,c, consider the Diophantine equation f(x+y+z) = xyz.  In other words, for fixed integers a,b,c we have the equation

If there exists a solution x,y,z to this equation, then z is a root of the quadratic

This can also be written in the form

Interestingly, we have z² = f(x+y)/a for a solution (x,y,z) if and only if (x,y) is a solution of f′(x+y) = xy.

As an example, consider the quadratic function f(s) = s² – s + 2, which has the discriminant .  The solutions can be characterized by the values of (x,y) for which z is an integer.  These smallest solutions can be arranged as shown below.

This tree can be extended infinitely in both directions.  Every pair of connected boxes contains four numbers, but one of them is duplicated, so there are three distinct numbers (except in the degenerate case of the [8,8] box), and these three numbers constitute a solution.

For another example, consider the quadratic function f(s) = s² – s + 6, which has the discriminant .  The smallest solutions can be arranged as shown below.

For some polynomials, such as f(s) = x² – s + c for c = +1, ±3, ±4, and +5 there are no solutions at all, but when there are solutions there are usually infinitely many, based on the tree construction.  However, for the quadratic f(s) = x² – s – 5 there is exactly one solution in positive numbers, namely (1,1,1), and the other root is –3, so we also have the solution (1,1,–3).

If we require the factors x,y,z to be primes, then equation (1) can be written in the form ξ(f(x)) = x, where x(n) signifies the sum of the prime factors of n.  As discussed in the note on Quadratic Sum of Prime Factors, the polynomial x² – x + 41 is interesting in this regard, because it happens that ξ(x² – x + 41) – x takes on the value –609 unusally often.  For these cases if we replace x with (x–609) we have ξ(x² + 1217x + 370313) = x.  The solutions of this equation are obviously just a subset of the solutions of the purely algebraic equation

The positive integer solutions of this equation evidently consist of four "trees", whose roots happen to involve some of the same integers.  The smallest solutions on each of these four trees are shown in the figures below.

The topology of these trees is interesting in its own right.  In general they follow the pattern shown below.

The only new information on the second row is the number c, and the only new information on the third row is the numbers d and e, and the only new information on the fourth row is the numbers f, g, h, and i.  Hence we could abbreviate the notation by just showing the new information on each level, as illustrated below.

Each of these letters represents two pairs of numbers, one of which consists of the given number joined with the number immediately above it.  The other pair consists of the given number joined with a number some levels higher.  To make this reference implicit in the arrangement of the tree, we can place the numbers on the left (or right) half of the nth level so that, for each number, the two reference indices back to the higher levels are the number of powers of 2 dividing the even numbers from 2 to 2n.  For example, the left half of the tree shown above on the 4th level (counting a,b as the 0th level) consists of the four numbers j, k, l, and m, and to these numbers we assign the pairs (16,14), (12,10), (8,6) and (4,2) respectively.  Next we replace each of these numbers with the greatest power of 2 dividing the number.  Thus we have the index pairs (4,1), (2,1), (3,1), and (2,1) respectively.  These indicies signify the number of levels up to the components of the two solution pairs for each number.  The number "j" has the indices (4,1), so the two solution pairs containing j (at this level) consist of joining j with the number 4 levels higher, and with the number 1 level higher.  Thus the two solution pairs corresponding to "j" in this tree are (j,a) and (j,f).  (Note that we choose either "a" or "b" at the zeroth level, depending on which half of the tree contains the reference number.)  Likewise the indices for "k" imply that the two solution pairs are (k,f) and (k,d).

The proof that the numbers can be arranged in this way is immediate, because on any given level, each pair consists of one "new" number (i.e., the new information on that level) and one number that has been carried down from the previous level.  Hence the two successors of this pair will each consist of one "new" number combined with each of the predecessor numbers.  So, we know one of the indices is 1, and the other index is one greater than the "old" index from the previous level.  So, if we encode the "old" indices on each level as the multiples of 4 in the range 2 to 2n, and the "new" indices as the remaining even numbers in that range, always pairing 4s with 4s–2, then the encoded indices for the successor are [2(4s), 2(4s)–2] and [2(4s–2), 2(4s–2)–2].

Another interesting way of arranging the solutions is in a pattern similar to multi-valued analytic surfaces in complex function theory.  This comes from the observation that each number which is part of a solution can be regarded at a "pole" in the surface of solutions, and it is the pivot point for a helical surface of infinitely many windings.  For a planar arrangement it is most natural to wrap these windings modulo 6 in a hexagonal pattern as shown below.  The solutions shown in this figure are for the polynomial f(q) = q² – q + 6.  We begin by placing the elements of the minimal solution (6,12,13) on one triangular element of the surface, and then we "continue" the surface by placing adjacent triangular elements with the same two numbers on the common edge.  Each triangular region represents a solution of f(x+y+z) = xyz.  Proceding clockwise around the "13" vertex, we place the solutions (13,6,29), (13,29,288), and so on.  Of course, beginning from (13,6,12) we can also proceed in the counter-clockwise direction, placing the solutions (13,12,101), (13,101,1074), and so on.  We can conceive of these infinite sequences of triangles as forming ascending and descending helical surfaces, winding around the "13" vertex.

Likewise there is an infinite helix of solutions surrounding every other vertex, so in a sense, the hexagons are multi-valued, and the sheaf that applies for any given evaluation depends on the path on the surface by which the location was reached.

The choice of the modulus six may seem arbitrary, but it is suggested by the fact that each solution contains three numbers, so there are three directions in which it can be modified, leading to the triangular representation where each edge corresponds to a transition to another neighboring solution.  Multi-valued surfaces of this kind (without the algebraic components) are discussed elsewhere in articles on “The No-Curvature Interpretation of General Relativity” and “The Tetratorus and Other Multi-Layered Polyhedra”.

Perhaps the most straight-forward way of arranging the solutions, and of preserving the metrical as well as the topological relations, is to regard the components of each solution as coordinates in a three-dimensional Cartesian coordinate system.  We can then connect "neighboring" solutions with line segments to show the connectedness and distance between solutions.  Since "neighbors" are defined as solutions differing by only one component, each connecting line will be parallel to one of the three axes.  The magnitudes of the components increase rapidly, so an alternative representation is to define the coordinates as the logarithms of the corresponding solution components.  Thus, for example, the solution (6,12,13) is mapped to the location (1.79, 2.48, 2.56).  Of course, there are six possible permutations of the components of any given solution, so each solution could be mapped to six different points.  However, beginning with a fixed permutation of the lowest order solution, (6,12,13), and generating all others by finding the neighbors that differ in only one component, we will be restricted to just a single point for each solution.  (This is because it isn't possible to go from (6,12,13) to (6,13,12), for example, by means of the "neighbor" operation.)  A plot of the first several levels of solutions for f(s) = s² – s + 6 in three-dimensional space is shown below.

The same solutions projected on the xy, xz, and yz planes are shown below.

Three oblique views of the same solutions, showing the truncation due to taking only one of the six possible permutations of each solution, are shown in the figures below.

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