Pythagoras, a Greek philosopher, around the 5th Century BC generalized the theorem which
states that in a right triangle the area of the square of the hypotenuse is the sum of the areas
of the squares of the other two sides.
Then Pierre de Fermat (1601-1665), a French lawyer and amateur mathematician, came
along and circa 1637 wrote in the margin of his personal copy of Bachet's translation of
Diophantus' Arithmetica. He wrote in Latin, "Cubum autem in duos cubos, aut
quadratoquadratum in duos quadratoquadratos & generaliter nullam in
infinitum ultra quadratum potestatem in duos eiusdem nominis fas est diuidere cuius rei
demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet." That is,
This result has come to be known as Fermat's Last Theorem (FLT). Andrew Wiles, a
professor at Princeton University, provided in two articles published in the May 1995 issue
of Annals of Mathematics.
Since 27 + 64 + 125 = 216 = 6^{3}, the Babylonian result obviously holds
when n = 3. Hence, the
answer to the above
existence question is in the affirmative. The
Babylonian result also holds when the three numbers were doubled, tripled, or multiplied by
any positive number. This generalized Babylonian result implies that if there are three
perfectly round balls with radiuses 3, 4, and 5, the total volume of the three balls is exactly
equal to that of another ball with radius equal to 6.
When there are three circles with radius x, y and z, and if the sum of the areas of the first
two circles with radiuses, x and y, equals that with radius z, Pythagoras's Theorem tells the
exact geometric procedure to get z: use the hypotenuse derived from a right triangle with
the two sides, x and y. However, in the three dimensional case, a geometric means or
procedure to get the radius of the ball whose volume is equal to the sum of the other three
balls has yet to be found.
3. What Is a Right Question?
If the intent is to extend the Babylonian triples to higher dimensions, once existence is
shown for n = 3, there is no need to stop here.
Consider a Diophantine equation which consists of finding k n^{th} powers
equal to an n^{th} power. Such an equation might be called an "n × k
equation." (This definition is somewhat different from the usual "m-n equation," and our
problem is a special case, i.e., n-1 Diophantine equation). Fermat's Last Theorem simply
states that no solution exists for n × 2, if n > 2. Our interest here is limited to the
case of n × n Diophantine equation. Solutions are known to exist for 2 × 2 (the
Babylonian triples), 3 × 3, 4 × 4, and 5 × 5. However, no known
solutions exist for n × n for n 6.
Cursory reading of a few books on the subject has not revealed any
conjecture or a theorem on the "even" or "balanced" case, where the number of summands
is equal to the power of z.
Specifically, I am interested in the following existence question:
Are there positive integers x_{1}, x_{2}, ..., x_{n}, z,
and n such that
x_{1}^{n} +
x_{2}^{n} + ...
x_{n}^{n} = z^{n} for all positive integer n >
2?
When n = 1, the above is trivial, and hence an alternative representation of the existence
question can be written as:
Conjecture 1: There exist positive integers x_{1},
x_{2}, ..., x_{n}, z, and n for the n × n Diophantine
equation
(1) x_{1}^{n} +
x_{2}^{n} + ...
x_{n}^{n} = z^{n} for all positive integer n
1.
The Babylonian triples suggest that an integer square can be expressed as the sum of two
integer squares. Intuitively, in the spirit of Fermat, the theorem, if
true, states that n^{th} power of a positive integer could be expressed as the
sum of n (rather than 2) n^{th} powers of some other smaller integers.
Davis
Wilson
has compiled a long list of the smallest n^{th}
powers which are the sums of distinct smaller n^{th} powers. For instance,
15^{4} = 4^{4} + 6^{4} + 8^{4} +
9^{4} + 14^{4}, and 12^{5} = 4^{5} +
5^{5} + 6^{5} + 7^{5} + 9^{5} +
11^{5}. As the powers increase, however, the number of terms of
n^{th} powers on the right hand side seems to increase much faster when the
smallest n^{th} powers are used. If integers are not limited to only
the smallest n^{th} powers or nondistinct integers are allowed, the
number of terms of n^{th} powers might be much smaller. Nevertheless, this
observation suggests that as the power increases, the chance for the existence theorem to
hold might decrease.
Is the existence theorem (or conjecture) in equation (1) true? If so, how do we
prove it? If not, why not? Answering this existence or nonexistence question might be a
real challenge.
Babylonians had many examples of the triples when n = 2. John Conway also gives another
example of the quadruple, (1, 6, 8, 9) when n = 3, and I am sure there are many others.
Noam Elkies also gave a monstrous example with nonnegative integers when n = 4, namely
the quintuple (0, 2682440, 15365639, 18796760, 20615673), and Roger Frye found a
smaller quintuple (0, 95800, 217519, 414560, 422481).[See Lopez-Ortiz and "Several
Years," Math. Comp. 51 (1988), 825-835)]. Although interesting, these examples involve
zero as a member of the quintuple.
Note that in the Babylonian example, you can add more integer variables. For instance,
from 3^{2} + 4^{2} = 5^{2}, 5^{2} +
12^{2} = 13^{2}, and 13^{2} + 84^{2} =
85^{2}, ..., we also get
3^{2} + 4^{2} + 12^{2} =
13^{2},
or
3^{2} + 4^{2} + 12^{2} +
84^{2} = 85^{2}.
Thus, Pythagorean N-tuple states that an odd integer square can be broken into (N - 2) even
integer squares and one odd square (Oliverio, 1996).
Moreover, by Fermat's Last Theorem, even though there exist no positive integers, x, y, and
z for which
x^{3} + y^{3} =
z^{3},
The quadruples (3,4,5,6) and (1,6,8,9) demonstrate that there exist positive integers, x, y, w,
and z for which
x^{3} + y^{3} + w^{3} =
z^{3}.
I suppose, for given n, adding more integer variables increases "freedom," and makes it
easier to find the modified theorem to hold. Thus, we should be able to find somewhat
simpler quintuples with all positive integers than the monstrous ones Elkies and Frye found
for n = 4.
Thus, another interesting conjecture is:
Conjecture 2:
Suppose there exist integers, x_{1}, x_{2}, ..., x_{k},
k, and n, 1 < k, for the n × k Diophantine equation
(2) x_{1}^{n} +
x_{1}^{n} + ... + x_{k}^{n} =
z^{n}.
Then a solution to equation (2) also exists for k + 1.
Note that when k = 1 and n = 3, the trivial Fermat equation,
x_{1}^{3} = z^{3} permits an easy solution,
x_{1} = z. In this case, adding another variable, x_{2}, yields:
x_{1}^{3} + x_{2}^{3} =
z^{3},
for which an integer solution does not exist by FLT. This explains the restriction, 1 < k < n,
in Conjecture 2. But why is this restriction necessary? When k = 2 and n = 3, we already
know that adding another variable suddenly permits a solution to Fermat's equation. Is there
periodicity here? Probably not. We have already seen that three fourth powers of positive
integers can sum to another. Moreover, it is known that four fifth powers of positive
integers can sum to another. Thus, periodicity is not likely to be present here, and that is the
reason for the above restriction, k > 1.
Shall we follow the footsteps of other mathematicians and prove the existence for each n?
or will there emerge another Andrew Wiles who will prove the general existence theorem or
non-existence theorem beyond some k? In any case, proving the existence conjecture for
each n is probably too time-consuming, and computers might help us prove existence up to
a very large number. However, a general proof might be needed, and the mathematical
machinery that has been developed so far might just be sufficient.
We should all congratulate Andrew Wiles wholeheartedly, and many others who have
contributed to various steps that led to the proof of the 350 year old puzzle. However, I
often
wonder how many pages God would have needed to prove FLT (with apologies to God for
presumptuously comparing divinity and humanity). Mathematicians in other inhabited
planets might have proved it in fewer lines. It is said "the argumentative defense of any
proposition is inversely proportional to the truth
contained." This adage seems to suggest a 200 page-long proof of Fermat's
Last Theorem might be a Pyrrhic victory. The two conjectures in this note may offer a new
challenge.