A Note on Fermat's Last Theorem—Was It a Right Question?

Carolan's concerto (sequenced by Barry Taylor)

1. Introduction

          An old Babylonian tablet (1900 - 1600 BC), shown on the right, contains the so-called Pythagorean Theorem, except that it predates Pythagoras by a millennium or more. A translation of another Babylonian tablet preserved in the British museum states (John Heise):

4 is the length and 5 the diagonal. What is the breadth? Its size is not known. 4 times 4 is 16. 5 times 5 is 25. You take 16 from 25 and there remains 9. What times what shall I take in order to get 9? 3 times 3 is 9. 3 is the breadth.

          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,

It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain.

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.


2. Fermat's Last Theorem—An Answer to a Wrong Question?

          It is important to note that since Pythagoras's Theorem holds for any right triangles, it has nothing to do with integers. That is, the length of the two sides need not be rational numbers. It is then clear that Fermat was looking for a generalization of the Babylonian result:

32 + 42 = 52.

Other sets of integers have since been found. (See e.g., van der Poorten, 1995)

x y z
3 4 5
5 12 13
7 24 25
9 40 41
... ... ...
x (x2 - 1)/2 (x2 + 1)/2
2ab (b2 - a2) (b2 + a2)

          In the penultimate line of the table, more sets of numbers can be generated using any rational numbers and letting y = (x2 - 1)/2, or (1 - x2)/2 and eliminating fractions by multiplying all three numbers by a suitable integer. Here, x can be either odd or even numbers. For instance, if x' = 4, then y' = 15/2 and z' = 17/2. Multiplying the triple by 2, we get (x,y,z) = (8, 15, 17). If 6 is used as the starting point, we get (x,y,z) = (12, 35, 37), etc. If a fraction like 8/7 is used, we first get (x',y',z') = (8/7, 15/98, 113/98), and then (x,y,z) = (112, 15, 113), etc.

          Pierre de Fermat asked whether the Babylonian result can hold for n 3. That is, he asked:

Are there positive integers x, y, z, and n > 2 such that

xn + yn = zn?

He conjectured that the answer was no. Mathematicians generally agree that Fermat did not have a proof.

I would argue that Fermat's Last Theorem was an answer to possibly a wrong question, if the intent was to extend the Babylonian triples to higher dimensions.

A Generalized Babylonian Result

          It is important to note that in the case of the Babylonian triples, the exponent was 2 and two integer variables (or sides) were used. Thus, if the intent was to extend the Babylonian result to higher dimensions, a more sensible starting point might be:

Are there positive integers, x, y, z, w and n = 3 such that

xn + yn + wn = zn?

During my brief stay in Ann Arbor, Jay Wilson stimulated my interest in FLT. Noticing the three consecutive numbers, 3, 4, and 5, in the Babylonian result, I was wondering if this sequence would extend to the three dimensional case. That is,

33 + 43 + 53 = 63?

Since 27 + 64 + 125 = 216 = 63, 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 nth powers equal to an nth 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 x1, x2, ..., xn, z, and n such that

          x1n + x2n + ... xnn = zn 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 x1, x2, ..., xn, z, and n for the n × n Diophantine equation

(1)           x1n + x2n + ... xnn = zn 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 nth power of a positive integer could be expressed as the sum of n (rather than 2) nth powers of some other smaller integers.

          Davis Wilson has compiled a long list of the smallest nth powers which are the sums of distinct smaller nth powers. For instance, 154 = 44 + 64 + 84 + 94 + 144, and 125 = 45 + 55 + 65 + 75 + 95 + 115. As the powers increase, however, the number of terms of nth powers on the right hand side seems to increase much faster when the smallest nth powers are used. If integers are not limited to only the smallest nth powers or nondistinct integers are allowed, the number of terms of nth 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 32 + 42 = 52, 52 + 122 = 132, and 132 + 842 = 852, ..., we also get

32 + 42 + 122 = 132,


32 + 42 + 122 + 842 = 852.

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

x3 + y3 = z3,

The quadruples (3,4,5,6) and (1,6,8,9) demonstrate that there exist positive integers, x, y, w, and z for which

x3 + y3 + w3 = z3.

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, x1, x2, ..., xk, k, and n, 1 < k, for the n × k Diophantine equation

(2)           x1n + x1n + ... + xkn = zn.

Then a solution to equation (2) also exists for k + 1.

Note that when k = 1 and n = 3, the trivial Fermat equation, x13 = z3 permits an easy solution, x1 = z. In this case, adding another variable, x2, yields:

x13 + x23 = z3,

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.

Amir D. Aczel, Fermat's Last Theorem: Unlocking the Secret of Ancient Mathematical Problem, Four Walls Eight Windows, New York, October 1996.
Alex Lopez-Ortiz, Fermat's Last Theorem, February 20, 1998.
Oliverio, P., "Self-Generating Quadruples and N-tuples," Fib. Quart. 34, 98-101, 1996.
Alf van der Poorten, Notes on Fermat's Last Theorem, Canandian Mathematical Society Series of Monographs and Advanced Texts, John Wiley & Sons, Inc., New York, 1995.
André Weil, Number Theory, An approach through history, From Hammurapi to Legendre, Birkhäuser, Boston, 1983.

I am grateful to David Rusin and S Butcher for helpful comments on an earlier version.
Timeline of Fermat's Last Theorem

October 12, 1998. © Kwan Choi.