Math Question - Square

Suppose the two different numbers be x and y respectively.


According to the given info. we have


I.x = y^2
II.y = x^2


Then we would have


x = y^2 = x^4,


x( x^3 - 1 ) = 0


x = 0 ( rej. as y = 0 as well ) or x^3 = 1


Symmtetrically, we have y^3 = 1


As this equation gives one real solution, it does not give distinct numbers of real soltion.


Hence, we conclude it has no real solution.


We thus consider the complex solution.


Obviously, the 1st and the 2nd roots of unity at degree 3 are the solution.


Hence, we have ( x, y ) = ( cis 2Pi/3, cis 4Pi/3 ) or ( cis 4Pi/3, cis2Pi/3 ).


Hope I can help you.

設x，y
x = y2
y = x2
x可以係1，y係開方1

i^2 = -1

(-i)^2 = (-i) * (-i) = i^2 = -1

設x，y
x = y2
y = x2
x可以係1，y係開方1
對嗎﹖