Fermat’s method of "descente inﬁnie" - Proof of Fermat’s Last Theorem for n=4
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Fermat proved the impossibility of ﬁnding positive in-
teger solutions of the equation x
by a method
which he claimed would ’enable extraordinary develop-
ments to be made in the theory of numbers’ , His
method was based upon the idea of using a supposed
solution to ﬁnd another solution, smaller than the ﬁrst.
Proceeding in this fashion he could therefore obtain an
inﬁnite series of smaller and smaller solutions. At this
point he concluded his proof as follows:
’This is, however, impossible because there cannot be
an inﬁnite series of numbers [positive integers] smaller
than any given [positive] integer we please. The margin
is too small to enable me to give the proof completely and
with all details .’
The margin referred to here was in Fermat’s copy of
the Arithmetica, an ancient Greek text written by Dio-
phantus of Alexandria. The problems given in this text
provided the stimulus for many of Fermat’s discoveries
in number theory. The equation x
problems concerning right-angled triangles
Fermat gave suﬃcient details of his proof to enable it
to be reconstructed. However, the purpose of this note
is not to reproduce the original proof faithfully but to
modify and simplify it whilst retaining the historical con-
nection with right-angled triangles.
Theorem: A right-angled triangle with rational sides
cannot have two sides each with a length equal to a square
or twice a square.
Proof. Suppose such a triangle did exist. By scaling
by a square number we can further assume the sides to
have integer lengths. Then, any prime factor p of two of
the lengths would be a factor of all three lengths. Fur-
thermore, if p were odd, then p
would be a factor of two
of the lengths and thus of all the lengths. By cancelling
down as necessary, we can therefore assume the lengths
to be coprime in pairs. Then, by Euclid’s formula, there
are coprime positive integers a and b, one odd and one
even, such that the triangle is as shown.
Now y and z are both odd and so neither can be twice
a square. If they were both squares then the triangle
shown below would be a new solution.
Otherwise, just one of y and z is a square and x = 2ab
is a square or twice a square. Then each of a and b is a
square or twice a square and one of the triangles shown
below would be a new solution.
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 be no
triangle with the required properties.