We want to solve the equation over the integers.
By writing we can factor on as
Using congruences modulo , one can show that both must be odd, and it can also be shown that and are relatively prime (if it were not the case, any divisor would have even norm, which is not possible).
Therefore, by unique factorization, and using that the only units (http://planetmath.org/UnitsOfQuadraticFields) on are , we have that each factor must be a cube.
So let us write
Then and . These two equations imply and thus , from where the only possible solutions are .
- 1 Esmonde, Ram Murty; Problems in Algebraic Number Theory. Springer.
|Date of creation||2013-03-22 14:52:05|
|Last modified on||2013-03-22 14:52:05|
|Last modified by||CWoo (3771)|
|Synonym||finding integer solutions to|