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§2. Fermat Descent and x + y = 1 127
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§2. Fermat Descent and x x x + y y y = 1
This section is a short detour away from the main purpose of the chapter and is not
used in the rest of the book. We wish to explain the context in which the Fermat
descent first arose.
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2
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We consider the equation x + y = z , where x, y,and z are integers without
common factors.
Step 1. We apply (3.1) of the Introduction to the primitive Pythagorean triple
2
2
2 2
2 2
2
x , y , z, where z = (x ) + (y ) giving a representation
2
2
2
2
2
2
x = m − n , y = 2mn, z = m + n .
2
2
2
Step 2. Now look at the relation x + n = m which says that n, x, m is a
2
2
Pythagorean triple which is again primitive, for otherwise x , y , z would not be
primitive. Applying (3.1) of the Introduction again, we can write
2
2
2
2
x = t − s , n = 2ts, m = t + s ,
where s and t have no common prime factor.
2
2
2
2
Step 3. Substituting into y , we obtain y = 2mn = 4st(s + t ).Since s and
t have no common prime factor, we deduce that both s and t are squares, that is,
2
2
s = a and t = b and hence
2
2 2
2 2
c = (a ) + (b )
for some c.
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2
2
Step 4. We started with a primitive z = x + y and we produced c = a + b 4
with the additional feature that z > c. To see that this inequality holds, we calculate
2 2
2
2 2
2
2
z = m + n = (t + s ) + 4t s
4 4 4 4 4 4 2 4
= (a + b ) + 4a b >(a + b ) = c .
This establishes the inequality which is the core of the Fermat descent procedure.
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(2.1) Theorem (Fermat, 1621). The equation x +y = t has no integral solutions
with both x and y nonzero. The only rational points on the Fermat curve
4 4
F 4 : x + y = 1
are the intersection points with the axes (±1, 0) and (0, ±1).
Proof. If the first equation had an integral solution x, y, t, then we could choose x,
y,and t strictly positive and consider a solution with t a minimum among all such
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solutions. By the Fermat descent procedure there would be a solution a + b = c 2
with c strictly smaller than t contradicting the minimal property of t. Hence there is
no integral solution.
For the second assertion, consider a rational solution and clear the denominators
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2
of x and y to obtain a relation between integers of the form u +v = t . By the first
result either u = 0or v = 0. This proves the theorem.