Therefore, = + and = – , substitute these two into the original equation, we get – 3 = ( + ) 3 + ( – ) 3 = 2 ( 2 + 3 2 ) Since 2 + 3 2 is odd and z 3 is even, u is even. Then we know that the greatest common divisor of 2 u and u 2 + 3 v 2 is either 1 (case A) or 3 (case B).
Proof for case A
2 = 3 2 + 3 2 = 3
There is a lemma states that if s is odd and if it satisfies an equation 3 = 2 + 3 2 then it can be written as s = 2 + 3 2 where e and f are two
coprime integers. Therefore, we get
= ( 2 − 9 2 ) = 3 ( 2 − 2 ) Because u is even and v is odd then e is even and f is odd. Substitute u into the equation for r 3 we get: 3 = 2 = 2 ( – 3 ) ( + 3 ) The factors of r 3 are coprime because 3 cannot divide e as a result, the three factors must individually equal cubes of smaller integers as follows
−2 = 3 − 3 = 3 + 3 = 3
Add the three equations up we get a solution 3 + 3 + 3 = 0 . Therefore, by the argument of infinite descent (the original equation can generate infinitely many of the same equations which makes the original equation unsolvable, the condition for this argument is that the numbers evolved cannot all be negative.), the original solution x , y , z is impossible.
Proof for case B
Because the greatest common divisor of 2 u and is 3, 3 divides u . U can be expressed as 3 w where w is a smaller integer. Substitute u =3 w into the equation for – z 3
− 3 = 6 (9 2 + 3 2 ) = 18 (3 2 + 2 )
Since v and w are coprime, then 18 w and 3 2 + 2 are coprime and hence they are each the cube of smaller integers, r and s
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