Sophie Germain: mathématicienne extraordinaire - Scripps College

The First Attempts 36
with Unique Factorization again in section 4.3.2. □
Returning to our proof we note that x 2 = 4ab(a 2 + b 2 ) is a square.
Recall that a and b are of opposite parity. Then ab is even and a 2 + b 2 is
odd.
By our lemma, since a, b, and a 2 +b 2 are relatively prime and ab(a 2 + b 2 )
is a square, then a is a square, b is a square, and (a 2 + b 2 ) is a square. We
write, a = X 2 and b = Y 2 . Then a 2 + b 2 = X 4 + Y 4 is a square, call it Z 2 .
We now have Z 2 = X 4 + Y 4 . Since a > b > 0, we know that Z 2 > 0.
Beginning this proof, we used the fact that z 4 0
was a square, ignoring
its quartic properties. That is, we said, for integers x 0 and y 0 , the integer
x 4 0 + y4 0 is a square which gives us new integers X and Y such that X4 + Y 4
is also a square. However, 0 < Z 2 = X 4 + Y 4 = a 2 + b 2 = s < s 2 + t 2 = z0 2,
that is, Z 2 < z0 2. So begining with a triple (x 0, y 0 , z0 2) such that x4 0 + y4 0 = (z2 0 )2 ,
we obtained a triple (X, Y, Z 2 ) such that X 4 + Y 4 = (Z 2 ) 2 and z 2 > Z 2 > 0.
So given any integer solution (x 0 , y 0 , z 0 ) with z 0 > 0, we get another
integer solution (x 1 , y 1 , z 1 ) such that z 0 > z 1 > 0. By the prinicple of infinite
descent, there can be no initial solution. Thus the sum of two fourth powers
can never be a square, much less a fourth power. Therefore, Fermat’s Last
Theorem for fourth powers (n = 4) is true. □
Now given any equation of the form x 4m + y 4m = z 4m , we may rewrite
it as
(x m ) 4 + (y m ) 4 = (z m ) 4 .