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

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

which is constructed from initial terms s_{0} = 0 and s_{1} = 1 , and each term is three times the previous term diminished by the term before that. The recursion equation is s_{n} = 3·s_{n1}  s_{n2} . For example, n = 5 , gives s_{5} = 55 = 3·21  8 . Note the identity s_{n} = F_{2n} 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, s_{4}·s_{1} = s_{3}·s_{2}  s_{2}·s_{1} = 8·3  3·1 = 21 . We solve this equation to get s_{4} = 21 . Similarly, we have also s_{5}·s_{1} = s_{3}·s_{3}  s_{2}·s_{2} = 8·8  3·3 = 55 which we solve to get s_{5} = 55 . These equations and related equations depend on triples of pairs of positive integers which form a pair of patterns

In other words, the following equations

are true for all positive n . These equations can be solved for s_{n} where n is greater than 2 given any values for s_{2} and s_{1} but s_{1} must be nonzero. 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 nonzero squares. It is known that every positive number is the sum of four squares, but not always four nonzero squares. The smallest such number is 4 since 4 = 1+1+1+1 is a sum of four nonzero 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 nonzero squares in more than one way. In this case, the smallest such number is 28 and we have the following three sums

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

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,

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 nonzero squares in three ways. It is 42. We have the following sums

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

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

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:
The sum of squares of each quadruple is equal to the sum of squares of the other two quadruples.
The product of the first quadruple is the product of the second quadruple minus the product of the third quadruple.
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 nonzero squares in more than two ways is 52 where

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

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

might suggest there is a simple arithmetic progression here. Does

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

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

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
a_{5}·a_{1}·a_{1}·a_{1} = a_{4}·a_{2}·a_{2}·a_{2}  a_{3}·a_{3}·a_{3}·a_{1} 
so if we know the value of a_{1},a_{2},a_{3},a_{4}, then the equation is linear in a_{5} and we solve for it (assuming nonzero a_{1}). Now, given a_{5}, the next equation is linear in a_{6}
a_{6}·a_{2}·a_{1}·a_{1} = a_{5}·a_{3}·a_{2}·a_{2}  a_{4}·a_{4}·a_{3}·a_{1} 
and we solve for it (assuming nonzero a_{1} and a_{2}). 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

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

which satisfies a_{n}·a_{n4} = a_{n1}·a_{n3}  a_{n2}·a_{n2} and many other equations. A closely similar example is

which satisfies a_{n}·a_{n4} = a_{n1}·a_{n3} + a_{n2}·a_{n2} and other equations. In fact, the odd indexed terms alternate in sign, and with the sign removed the resulting positive sequence

satisfies the same recursion equation. It is called the Somos4 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 4042) 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 F_{n} = F_{n1} + F_{n2} that defines the Fibonacci sequence or the one for the test sequences introduced in Section 1, that naturally produce integer sequences, the Somos4 recursion
a_{n} = (a_{n1}·a_{n3} + a_{n2}·a_{n2})/a_{n4} 
is rational. This is indeed wonderful that the sequence that starts with
We show inductively that if a_{n4}, a_{n3}, ..., a_{n+3} are integers, then so is a_{n+4} and hence every a_{i}. This is of course true for
(*)  a, b, c, 0, c^{2}/a, c^{3}/ab, c^{3}/a^{2}. 
Which leads to
a_{n}·a_{n+4} = (ac + b^{2})·c^{5}/(a^{3}b^{2}) = 0 (mod a_{n}) 
meaning that a_{n} divides a_{n}·a_{n+4}, or, in other words, that a_{n+4} is an integer.
(Note that the terms in (*) involve division by a and b modulo a_{n}. The operation is legitimate as easily proven by another induction from the observation that any prime factor of a_{n} and ac is aslo a factor of b, and vice versa. Induction then used to show that any four successive terms of the sequence {a_{i}} are pairwise mutually prime. In particular neither
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
 D. Gale, Tracking the Automatic Ant, SpringerVerlag, 1998
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