Step into the Elliptic Realm
by Michael Somos
29 Jan 2000

"A journey of a thousand miles begins with a single step."
-- Lao Tzu

"All is number."
-- Pythagoras

This is a brief report about my ongoing research in mathematics. In order to read and understand this you should have a knowledge of integer arithmetic and solvability of linear equations in one unknown, a familiarity with a little bit of algebraic notation, plus some curiosity about numbers and their interconnections. The goal is to discover a class of functions which generalize trigonometric functions using only the simplest tools possible. These new functions are a part of what I have figuratively called "the Elliptic Realm" after a usage of this term by Harris Hancock.

Section 1. Functions, Tables, and Sequence Construction.

In this context, a function will be specified by giving a table of values of the function at equally spaced points usually centered at origin zero. For example, the square function is given by the table

 
n: -5 -4 -3 -2 -1 0 1 2 3 4 5 ...
n·n: 25 16 9 4 1 0 1 4 9 16 25 ...

where the input values are spaced a unit distance from each other starting from zero. The table of values forms a sequence. We will be searching for sequences which can be calculated with just a few simple arithmetic operations and with some nice arithmetic properties in order to test out our ideas with.

A good example of such a test sequence is in the following table

 
n: 0 1 2 3 4 5 6 7 8 9 10 ...
sn: 0 1 3 8 21 55 144 377 987 2584 6765 ...

which is constructed from initial terms s0 = 0 and s1 = 1 , and each term is three times the previous term diminished by the term before that. The recursion equation is sn = 3·sn-1 - sn-2 . For example, n = 5 , gives s5 = 55 = 3·21 - 8 . Note the identity sn = F2n where F denotes the Fibonacci sequence. We will call this sequence Fib2 for reference later. Note that any integer could have been used instead of 3 to construct a test sequence.

There are a few other ways to generate the terms of this sequence. For example, s4·s1 = s3·s2 - s2·s1 = 8·3 - 3·1 = 21 . We solve this equation to get s4 = 21 . Similarly, we have also s5·s1 = s3·s3 - s2·s2 = 8·8 - 3·3 = 55 which we solve to get s5 = 55 . These equations and related equations depend on triples of pairs of positive integers which form a pair of patterns

 
[4,1] [3,2] [2,1] and [3,1] [2,2] [1,1]
[6,1] [4,3] [3,2] and [5,1] [3,3] [2,2]
... ... ...   ... ... ...
[2n+2,1] [n+2,n+1] [n+1,n] and [2n+1,1] [n+1,n+1] [n,n]

In other words, the following equations

 
s2n+2·s1 = sn+2·sn+1 - sn+1·sn , and
s2n+1·s1 = sn+1·sn+1 - sn·sn ,

are true for all positive n . These equations can be solved for sn where n is greater than 2 given any values for s2 and s1 but s1 must be non-zero. These sequences are named Chebyshev polynomials and can be used to construct sine tables using only simple arithmetic. However, we want to go beyond these test sequences.

Section 2. Sums of Four Squares and Elliptic Sequences.

To discover some new sequences we need a starting point. There are many possible starting points. I have come across what I think is the easiest so far. It begins with the sum of four non-zero squares. It is known that every positive number is the sum of four squares, but not always four non-zero squares. The smallest such number is 4 since 4 = 1+1+1+1 is a sum of four non-zero squares. Next on the list is 7 = 4+1+1+1 and then 10 = 4+4+1+1 . We now require the number to be a sum of four non-zero squares in more than one way. In this case, the smallest such number is 28 and we have the following three sums

 
 [5,1,1,1] [4,2,2,2] [3,3,3,1]
28= 25+1+1+1 = 16+4+4+4 = 9+9+9+1

where recorded above each square is its square root. The three square root quadruples have many arithmetic properties. For example, if we take the product of the four numbers of each quadruple we get

 
[5,1,1,1] [4,2,2,2] [3,3,3,1]
5 = 5·1·1·1 32 = 4·2·2·2 27 = 3·3·3·1

and it so happens that 5 = 32 - 27 . If we use each number of each quadruple as an index into the Fib2 test sequence mentioned earlier,

 
[5,1,1,1] [4,2,2,2] [3,3,3,1]
55 = 55·1·1·1 567 = 21·3·3·3 512 = 8·8·8·1

the calculated product of corresponding terms gives 55 = 567 - 512 . These quadruples have unexpected arithmetic properties when indexed into test sequences. Is this just unexplained coincidence? Not really.

To see why, find the next smallest number which is the sum of four non-zero squares in three ways. It is 42. We have the following sums

 
 [6,2,1,1] [5,3,2,2] [4,4,3,1]
42= 36+4+1+1 = 25+9+4+4 = 16+16+9+1

and again the square root quadruples are recorded above. As might be expected, we get the equality of difference of products as given by

 
[6,2,1,1] [5,3,2,2] [4,4,3,1]
12 = 6·2·1·1 60 = 5·3·2·2 48 = 4·4·3·1

and 12 = 60 - 48 . Also, taking the product of corresponding indexed terms of the Fib2 test sequence gives the expected result

 
[6,2,1,1] [5,3,2,2] [4,4,3,1]
432 = 144·3·1·1 3960 = 55·8·3·3 3528 = 21·21·8·1

which is 432 = 3960 - 3528 . It is time to summarize our findings. We have found some triples of quadruples of positive integers with the following properties:

  1. The sum of squares of each quadruple is equal to the sum of squares of the other two quadruples.

  2. The product of the first quadruple is the product of the second quadruple minus the product of the third quadruple.

  3. The second property holds if we use the product of the corresponding indexed terms of test sequences of numbers.

A reasonable goal is now to find all such triples and all the sequences of numbers related to them. The next sum of four non-zero squares in more than two ways is 52 where

 
 [7,1,1,1] [5,5,1,1] [5,3,3,3] [4,4,4,2]
52= 49+1+1+1 = 25+25+1+1 = 25+9+9+9 = 16+16+16+4

and so we seek the correct triple of quadruples. In this case it is

 
[7,1,1,1] [5,3,3,3] [4,4,4,2]
7 = 7·1·1·1 135 = 5·3·3·3 128 = 4·4·4·2

and 7 = 135 - 128 . What about the general case? There are a few clues here. For example, the first and third triples found compared

 
[5,1,1,1] [4,2,2,2] [3,3,3,1]
[7,1,1,1] [5,3,3,3] [4,4,4,2]

might suggest there is a simple arithmetic progression here. Does

 
[9,1,1,1] [6,4,4,4] [5,5,5,3]

give us another? Yes, it does, and the general pattern is given by

 
[2n+1,1,1] [n+2,n,n,n] [n+1,n+1,n+1,n-1]

This gives us one infinite set of triples. Are there any others? Yes. For example, starting from the second triple we get a pattern

 
[6,2,1,1] [5,3,2,2] [4,4,3,1]
[8,2,1,1] [6,4,3,3] [5,5,4,2]
[10,2,1,1] [7,5,4,4] [6,6,5,3]
... ... ...
[2n+2,2,1,1] [n+3,n+1,n,n] [n+2,n+2,n+1,n-1]

which is similar to the first pattern. From just these two patterns we can construct sequences in a very simple way.

Section 3. Constructing Elliptic Sequences.

To start the process we need the value of four initial sequence terms. The reason is that the first triple equation for a sequence is

  a5·a1·a1·a1 = a4·a2·a2·a2 - a3·a3·a3·a1

so if we know the value of a1,a2,a3,a4, then the equation is linear in a5 and we solve for it (assuming non-zero a1). Now, given a5, the next equation is linear in a6

  a6·a2·a1·a1 = a5·a3·a2·a2 - a4·a4·a3·a1

and we solve for it (assuming non-zero a1 and a2). So, given the equations from the two patterns we can solve for all the rest of the terms of the sequence. This gives a construction of new sequences, but what about other patterns? For example, starting from the first and the second triple leads to the pattern

 
[5,1,1,1] [4,2,2,2] [3,3,3,1]
[6,2,1,1] [5,3,2,2] [4,4,3,1]
[7,3,1,1] [6,4,2,2] [5,5,3,1]
... ... ...
[n,n-4,1,1] [n-1,n-3,2,2] [n-2,n-2,3,1]

Do the sequences fit into the other patterns? The answer is yes, but the proof is not obvious and requires work. What do non-trivial examples of these new sequences look like? A simple example is

 
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
an: 0 1 1 1 -1 -2 -3 -1 7 11 20 -19 -87 -191 -197 ...

which satisfies an·an-4 = an-1·an-3 - an-2·an-2 and many other equations. A closely similar example is

 
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
an: 0 1 1 -1 1 2 -1 -3 -5 7 -4 -23 29 59 129 ...

which satisfies an·an-4 = an-1·an-3 + an-2·an-2 and other equations. In fact, the odd indexed terms alternate in sign, and with the sign removed the resulting positive sequence

 
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
an: 1 1 1 1 2 3 7 23 59 314 1529 8209 83313 ...

satisfies the same recursion equation. It is called the Somos-4 sequence - the simplest member of a large family of sequences.

All this and more is part of the Elliptic Realm. These sequences are related to Jacobi theta functions and Weierstrass sigma functions discovered in the nineteenth century. Among several names they were called "elliptic divisibility sequences" by Morgan Ward in 1948. I have demonstrated that they can easily be discovered and constructed using simple tools, after a suitable starting point is chosen.

The first popular exposition of the Somos sequences appeared in D. Gale's column in the Math. Intelligencer in 1991 (13, pp 40-42) and was later reprinted in 1998 as the first chapter of his book, Tracking the Automatic Ant. In the book, Gale remarks that "Somos actually discovered his sequences 14 years ago but did not succeed in capturing attention of the mathematical community until the summer of 1989."

Unlike the recursion Fn = Fn-1 + Fn-2 that defines the Fibonacci sequence or the one for the test sequences introduced in Section 1, that naturally produce integer sequences, the Somos-4 recursion

  an = (an-1·an-3 + an-2·an-2)/an-4

is rational. This is indeed wonderful that the sequence that starts with a0 = a1 = a2 = a3 = 1 consists entirely of positive integers. Gale gives a variant of a proof due to George Bergman.

We show inductively that if an-4, an-3, ..., an+3 are integers, then so is an+4 and hence every ai. This is of course true for n = 4. Let an-3 = a, an-2 = b, and an-1 = c. an·an-4 = ac + b2. So that an divides ac + b2. Modulo an, the terms of the sequence (starting with an-3) become :

(*) a, b, c, 0, c2/a, c3/ab, c3/a2.

Which leads to

  an·an+4 = (ac + b2)·c5/(a3b2) = 0 (mod an)

meaning that an divides an·an+4, or, in other words, that an+4 is an integer.

(Note that the terms in (*) involve division by a and b modulo an. The operation is legitimate as easily proven by another induction from the observation that any prime factor of an and ac is aslo a factor of b, and vice versa. Induction then used to show that any four successive terms of the sequence {ai} are pairwise mutually prime. In particular neither an-3 = a nor an-2 = b may shared prime factors with an.)

For additional examples of rational recursions that generate integer sequences see D.Gale's book or follow the links at the end of M. Somos' essay.

(The above proof works for a much simpler sequence for which there is an online calculator.)

References

  1. D. Gale, Tracking the Automatic Ant, Springer-Verlag, 1998


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