23/11/13 12:21:36.31 KTxljM+6.net
It is inconsistent since z^(n-1)≡0 (mod x+y) holds contrary to the expression (3). Thereby, there are no solutions to the equation (1) when n≡0 (mod x+y) holds.
In the second place, we consider the case where n≢0 (mod x+y) holds. Since a=農(i=1)^n▒(-1)^(i-1) x^(n-i) y^(i-1) holds,
a≡nx^(n-1)≢0 (mod x+y) …(4)
holds. By the equation (2),
(x+y)a=((x+y)/b)^n
ab^n=(x+y)^(n-1)
holds. By the expression (4), a=1 holds. In this case, by the equations (1) and (2),
x^n+y^n=x+y
holds. x=1 and y=1 is the only positive solution when n≧3 holds. Then z is not an integer since z^n=2 holds.