What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

§3. PYTHAGOREAN NUMBERS AND FERMAT’S LAST THEOREM

An interesting question in number theory is connected with the Pythagorean theorem. The Greeks knew that a triangle with sides 3, 4, 5 is a right triangle. This suggests the general question: What other right triangles have sides whose lengths are integral multiples of a unit length? The Pythagorean theorem is expressed algebraically by the equation

where a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse. The problem of finding all right triangles with sides of integral length is thus equivalent to the problem of finding all integer solutions (a, b, c) of equation (1). Any such triple of numbers is called a Pythagorean number triple.

The problem of finding all Pythagorean number triples can be solved very simply. If a, b and c form a Pythagorean number triple, so that a² + b² = c², then we put, for abbreviation, a/c = x, b/c = y. x and y are rational numbers for which x² + y² = 1. We then have y² = (1 – x)(1 + x), or y/(1 + x) = (1 – x)/y. The common value of the two sides of this equation is a number t which is expressible as the quotient of two integers, u/v. We can now write y = t(1 + x) and (1 – x) = ty, or

tx – y = –t, x + ty = 1.

From these simultaneous equations we find immediately that

Substituting for x, y and t, we have

Therefore

for some rational factor of proportionality r. This shows that if (a, b, c) is a Pythagorean number triple, then a, b, c are proportional to v² – u², 2uv, u² + v², respectively. Conversely, it is easy to see that any triple (a, b, c) defined by (2) is a Pythagorean triple, for from (2) we obtain

a² = (u⁴ – 2u²v² + v⁴)r²

b² = (4u²v²)r²,

c² = (u⁴ + 2u²v² + v⁴)r²,

so that a² + b² = c².

This result may be simplified somewhat. From any Pythagorean number triple (a, b, c) we may derive infinitely many other Pythagorean triples (sa, sb, sc) for any positive integer s. Thus, from (3, 4, 5) we obtain (6, 8, 10), (9, 12, 15), etc. Such triples are not essentially distinct, since they correspond to similar right triangles. We shall therefore define a primitive Pythagorean number triple to be one where a, b, and c have no common factor. It can then be shown that the formulas

a = v² – u²,

b = 2uv,

c = u² + v²,

for any positive integers u and v with v > u, where u and v have no common factor and are not both odd, yield all primitive Pythagorean number triples.

*Exercise: Prove the last statement.

As examples of primitive Pythagorean number triples we have u = 2, v = 1: (3,4,5), u = 3, v = 2: (5, 12, 13), u = 4, v = 3: (7, 24, 25), · · ·, u = 10, v = 7: (51, 140, 149), etc.

This result concerning Pythagorean numbers naturally raises the question as to whether integers a, b, c can be found for which a³ + b³ = c³ or a⁴ + b⁴ = c⁴, or, in general, whether, for a given positive integral exponent n > 2, the equation

can be solved with positive integers a, b, c. An answer was provided by Fermat in a spectacular way. Fermat had studied the work of Diophantus, the ancient contributor to number theory, and was accustomed to making comments in the margin of his copy. Although he stated many theorems there without bothering to give proofs, all of them have subsequently been proved, with but one significant exception. While commenting on Pythagorean numbers, Fermat stated that the equation (3) is not solvable in integers for any n > 2, but that the elegant proof which he had found was unfortunately too long for the margin in which he was writing.

Fermat’s general statement has never been proved true or false, despite the efforts of some of the greatest mathematicians since his time. The theorem has indeed been proved for many values of n, in particular, for all n < 619, but not for all n, although no counterexample has ever been produced. Although the theorem itself is not so important mathematically, attempts to prove it have given rise to many important investigations in number theory. The problem has also aroused much interest in non-mathematical circles, due in part to a prize of 100,000 marks offered to the person who should first give a solution and held in trust at the Royal Academy at Göttingen. Until the post-war German inflation wiped out the monetary value of this prize, a great number of incorrect “solutions” was presented each year to the trustees. Even serious mathematicians sometimes deceived themselves into handing in or publishing proofs which collapsed after some superficial mistake was discovered. General interest in the question seems to have abated since the devaluation of the mark,.though from time to time there is an announcement in the press that the problem has been solved by some hitherto unknown genius.