Part 4:  Higher-Order Solutions

 

Solution Triples:  A set of three distinct integers {M,N,T} such that Mx(M) = Nx(N) = Tx(T) will be called a solution triple.  We can partition three such integers into seven co-prime factors a,b,..,g as illustrated in Figure 1a.

 

 

Clearly, each of the pairs {M,N}, {M,T}, and {N,T} is a solution pair.  Considering the pair {M,N} we note that gcd(M,N) = eg, so the solution kernel of this pair is ad,bf.  By Proposition 2 we have

 

 

Expanding the left side by the additive property gives

 

 

Similarly the solution pair {N,T} leads to the condition

 

 

The difference between these last two equations gives a necessary and sufficient condition for {M,N,T} to be a solution triple:

 

 

One way of searching for such triples is to generate a list of solution (pair) kernels and then examine these pairs two at a time to determine if they satisfy (4).  However, each set of solution pairs actually corresponds to two possible solution triples.  Consider the three pairs  {M,N}, {M,T}, and {N,T} illustrated in Figure 1a.  The kernels corresponding to these pairs are {ad,bf}, {ae,cf},{be,cd}.  Now consider the three pairs {M',N'}, {M',T'}, and {N',T'} illustrated in Figure 1b.  The kernels of these pairs are {bf,ad}, {cf,ae}, and {cd,be}, which are identical to the pairwise kernels for {M,N,T}, in spite of the fact that {M,N,T} and {M',N',T'} are distinct.  This occurs because there is a permutation of the components a,b,..,f  that leaves the pairwise kernels unchanged (except that the components are transposed).  Specifically, the permutation consisting of the three transpositions

 

 

produces a new triple with the same pairwise kernels. We refer to these two triples as duals of each other.

 

Given two solution kernels {u,v} and {r,s}, we could make the assignments

 

 

and then compute the individual components as follows

 

 

If these components satisfy (4) then we call the integers a,b,..,f a triple kernel.  We can then use (3) to compute x(g).  For any g having this value of x(g) the integers M = adeg, N = bfeg, T = cdfg constitute a solution triple. 

 

On the other hand, given the same solution pairs {u,v} and {r,s}, we could equally well make the assignments

 

 

in which case the individual components are given by

 

 

If these components satisfy (4) then the integers M',N',T' as defined in Figure 1b are a solution triple.  As an example, consider the two solution pairs {12,11} and {22,19}.  If we make the assignments

 

 

then we have the components

 

 

which gives the three integers (according to Figure 1a)

 

 

These numbers do not constitute a solution triple, since the components fail to satisfy equation (4).  However, the "dual" of this triple gives the components

 

 

yielding the three integers

 

 

which do satisfy the requirements of a solution triple with g defined such that x(g) = 18.

 

Evidently any number of triple kernels can be found in this way.  Table 4 contains those that can be produced from the double kernels of Table 1, but evaluating only one of the two possible permutations in each case, so the list is incomplete. The smallest solution triple is

 

 

which is based on the kernel

 

 

taking g = 21.

 

We observe that most of the kernels listed in Table 4 have at most only one of a,b,c > 1.  Also, it appears that the six integers

 

 

constitute a triple kernel if and only if  u,v  and  u,w (allowing transpositions) are double kernels with q(u,v) = q(u,w).

 

Higher Order Solution Sets:  Similarly we can determine sets of 4 integers A,B,C,D such that

 

Ax(A) = Bx(B) = Cx(C) = Dx(D)

 

and Table 4 provides several examples. One such 4-solution consists of the integers

 

 

which satisfy the equality

 

 

Table 4 also gives examples of 5-solutions.  If we define

 

 

then the integers

 

 

for any k such that x(k) = 52 constitute a 5-solution.  This is equivalent to the statement that the set of primes

 

S = {2,2,3,5,11,17,19,29,59,71,89}

 

contains 5 distinct subsets s1, s2, ..., s5 such that

 

 

where

 

 

If we add the prime 47 and another 5 to S, then the 52 in the above equation vanishes.  However, the primes 5 and 47 are, in a sense, inert elements of the set.  (We consider later whether there are such sets with no inert elements.)

 

We might remove the restriction that the elements of S be primes, and instead search for sets of general integers such that

 

 

A further generalization is to allow the constant to be negative. The subsets si may or may not be disjoint and covering (as they are in the preceding example).

 

Finding Solution Sets of Order k:  Proposition 4 stated that  {m,n} constitutes a solution kernel if and only if the quantity

 

 

is a positive integer.  This can be rewritten in the form

 

 

which is true if and only if

 

 

(because we can expand the x functions and subtract x(q) from both sides).  It follows that for any two integers M and N such that M + x(M) = N + x(N)  and  gcd(M,N) = q,  where q is any positive integer, the integers M/q and N/q constitute a solution kernel.

 

This can be immediately generalized to solution sets of k elements.  For any positive integer c let c(c) denote the number of integers n such that  n + x(n) = c.  (This is called the valence of the function G.)  Also, let t(c) denote c minus the sum of the values x(n) over this set of  n values.  Then, given this set of k integers n1, n2, ..., nk such that

 

 

we can construct a set of integers N1, N2, ..,Nk such that

 

 

by taking

 

where the quantities uj and s are defined by

 

 

and x-1 signifies the inverse x function.  Thus, the right-hand factor in the expression for Nj is an integer b such x(b) = t(c) + x(s).  (In general the function x-1 is multi-valued.)  This requires that t(c) + x(s) be positive.

 

Thus, every integer c with c(c) = k and  t(c) + x(s) > 0  corresponds to a solution set of k elements.  It is conjectured that the converse is also true, i.e., that every solution set of k elements is associated with an integer c for which c(c) is equal to (or greater than) k.  The values of x(c) and t(c) and the ni for each integer c from 1 to 500 are listed in Table 5.

 

It's useful to consider an example.  The first "triple" in Table 4 is defined by the values

 

 

This gives the family of solution triples

 

 

where g = x-1(165).  This same triple is represented in Table 5 where, for c = 216, we find

 

 

which gives

 

 

and

s   =   gcd(38000, 39000, 37050)   =   50

 

Noting that x(s) = 12, we see that this gives the same family of solution triples as found in the earlier table. 

 

Discussion:  We could simply have factored  s  out of the three integers (n1n2), (n1n3), (n2n3)  and asserted the solution triples  {760q, 780q, 741q}  where now  q = 50 x-1(153).  These values of q are a subset of the values of  g = x-1(165).  However, they do not constitute the complete set, because there are integers g for which x(g) = 165 but which are not divisible by 50.  For example, g = (163)(2).  This arises because of the non-uniqueness of the inverse x function.

 

Very High Order Solution Sets:  Table 7 gives the first few values of c for which c(c) equals 1, 2, ..., 24.  Presumably, for any given positive integer k there are infinitely many c such that c(c) = k.  The smallest integer c such that c(c) = 12 is 14399.  This leads to the 12-kernal with the values of ni shown below:

 

 

In this case c = 14399, the sum of the x(ni) is 13400, and the ni have no common factor, so b is any integer such that x(b) = 999.

 

The next smallest integer c for which c(c) = 12 is 15119.  The corresponding values of ni are as follows

 

 

In this case the sum of the x(ni) is greater than c, and the ni have no common factor, so we have x(b) = -2759, which is not satisfied for any integer b.

 

To give another example of a very high order solution set, we note that c(720719) = 24, which implies the existence of the 24-kernel presented in the following table.

 

 

This table shows that t(720729) = 219745 and there are no common factors in the ni.  Therefore, we can construct a family of 24-solutions using the formulas given above.  The number of solutions in this family is equal to the number of prime partitions of 219745.

 

Each value of ni in this 24-solution is of the form piqi where pi and qi are primes such that

 

 

An efficient way of searching for such solutions is to observe that we can add 1 to both sides and factor the resulting equation as follows:

 

 

Therefore, we need only determine the two-part multiplicative partitions ab = c + 1 such that a-1 and b-1 are primes.  In the present case we have c = 720719, so we need to examine the factors of

 

c + 1  =  720720  =  (2)4(3)2(5)(7)(11)(13)

 

Cunningham Chains:  The is an obvious tendancy for the ni factors to be primes of the form p = 2q+1 where q is also a prime.  Sequences of primes with this property are called Cunningham chains.  The first two Cunningham chains of unusual length are

 

 

If N = rs for primes r and s then N + x(N) = rs + r + s.  Now, if M = pq where p = 2r + 1 and q = (s-1)/2 (both primes), then we have

 

 

Thus, by forming a sequence of integers given by multiplying the elements of a Cunningham chain in increasing order times the elements of a chain in decreasing order gives integers with the same value of N + x(N).  For example, the smallest set of 11 integers with equal G functions is

 

 

In this case c = 4319 so x(b) = -159, which means that this set cannot be used to construct an 11-kernel.  However, this clearly illustrates how Cunningham chains can be used to produce solution sets.  We can also explain several of the 3-kernels by considering the convoluted chains

 

 

It is conjectured, but not proven, that there exist Cunningham chains of any finite length, which would imply the existence of solution sets of arbitrary size.

 

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