F ermat’s metho d of "descen te inﬁnie" - Pro of of F ermat’s Last Theorem for n=4

Stan Dolan

126A Harp enden R o ad, St. Alb ans

In tro duction

F ermat pro ved the imp ossibility of ﬁnding p ositiv e in-

teger solutions of the equation x

4

+ y

4

= z

4

b y a metho d

whic h he claimed would ’enable extraordinary dev elop-

men ts to b e made in the theory of n umbers’ [1], His

metho d was based up on the idea of using a supposed

solution to ﬁnd another solution, smaller than the ﬁrst.

Pro ceeding in this fashion he could therefore obtain an

inﬁnite series of smaller and smaller solutions. A t this

p oin t he concluded his pro of as follows:

’This is, however, imp ossible b e c ause ther e c annot b e

an inﬁnite series of numb ers [p ositive inte gers] smal ler

than any given [p ositive] inte ger we ple ase. The mar gin

is to o smal l to enable me to give the pr o of c ompletely and

with al l details [1].’

The margin referred to here was in F ermat’s copy of

the Arithmetica, an ancient Greek text written by Dio-

phan tus of Alexandria. The problems given in this text

pro vided the stimulus for many of F ermat’s disco veries

in n umber theory . The equation x

4

+ y

4

= z

4

arose from

problems concerning right-angled triangles

F ermat’s pro of

F ermat ga ve suﬃcien t details of his proof to enable it

to b e reconstructed. Ho wev er, the purp ose of this note

is not to repro duce the original proof faithfully but to

mo dify and simplify it whilst retaining the historical con-

nection with right-angled triangles.

Theorem: A right-angled triangle with rational sides

cannot ha ve t wo sides eac h with a length equal to a square

or twice a square.

Pro of. Supp ose such a triangle did exist. By scaling

b y a square num b er w e can further assume the sides to

ha ve integer lengths. Then, any prime factor p of tw o of

the lengths w ould b e a factor of all three lengths. F ur-

thermore, if p were o dd, then p

2

w ould b e a factor of tw o

of the lengths and thus of all the lengths. By cancelling

do wn as necessary , we can therefore assume the lengths

to b e coprime in pairs. Then, b y Euclid’s formula, there

are coprime p ositive in tegers a and b, one o dd and one

ev en, suc h that the triangle is as sho wn.

FIG. 1:

No w y and z are b oth o dd and so neither can b e t wice

a square. If they w ere both squares then the triangle

sho wn b elo w w ould be a new solution.

FIG. 2:

Otherwise, just one of y and z is a square and x = 2 ab

is a square or t wice a square. Then each of a and b is a

square or twice a square and one of the triangles shown

b elo w would b e a new solution.

FIG. 3:

In all cases, the new solution has a smaller hypotenuse

than the original triangle. Since there cannot be an in-

ﬁnite series of decreasing hypotenuses, there can b e no

triangle with the required prop erties.