Step into the Elliptic Realm
by Michael Somos
23 Mar 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
curiousity 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 ...
s(n): 0 1 3 8 21 55 144 377 987 2584 6765 ...
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(n-1) - s(n-2) .
For example, n = 5 , gives s(5) = 55 = 3*21 - 8 . Note the identity
s(n) = F(2*n) 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 three 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
[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
s(2*n+2)*s(1) = s(n+2)*s(n+1) - s(n+1)*s(n) , and
s(2*n+1)*s(1) = s(n+1)*s(n+1) - s(n)*s(n) ,
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 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
[2*n+1,1,1,1] [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]
... ... ...
[2*n+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
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 non-zero 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 non-zero 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
[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 ...
a(n): 0 1 1 1 -1 -2 -3 -1 7 11 20 -19 -87 -191 -197 ...
which satisfies a(n)*a(n-4) = a(n-1)*a(n-3) - a(n-2)*a(n-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 ...
a(n): 0 1 1 -1 1 2 -1 -3 -5 7 -4 -23 29 59 129 ...
which satisfies a(n)*a(n-4) = a(n-1)*a(n-3) + a(n-2)*a(n-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 ...
a(n): 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.