(In reply to

re: Oops! Solution by reading the problem. by Daniel)

Daniel:

Hmm. Right you are! I just substituted in N = +/- i, and I see that you are correct. They are both solutions of the equation. I wonder where I went wrong.

Also, once again,

^{5}√(N^{3} + 2N) = ^{3}√(N^{5} - 2N) = N

I just worked out why.

If ^{3}√(N^{5} - 2N) = N

then **(N**^{5} - 2N) = **N**^{3} by cubing each side

then **N**^{5} = **N**^{3} **+ 2N **by rearranging terms

and **N = **^{5}√(N^{3} + 2N) by taking the fifth root of each side

so ^{5}√(N^{3} + 2N) = ^{3}√(N^{5} - 2N) = N

I also just noticed that if

If ^{3}√(N^{5} - 2N) = N

then **(N**^{5} - 2N) = **N**^{3} then **(N**^{5} - **N**^{3}^{ }**-**^{ }**2N) = **^{ }0

and N ( N^2 + 1)(N^2 - 2) = 0

and N is 0 or sqrt(2) or -sqrt(2) or i or -i

I am not claiming these are the only solutions to the original equation

^{ }