Obviously N = 0 is one solution

For other solutions (N <> 0), raise each side to the 15th power

Then (N^3 + 2N)^3 = (N^5 -2N)^5

N^3 * (N^2 + 2)^3 = N^5 * (N^4 - 2)^5

Dividing by N^3

(N^2 + 2)^3 = N^2 * (N^4 - 2)^5

Let a = N^2

(a + 2)^3 = a*(a^2 -2)^5

a > 0 (because it equals N^2), so the left hand side (lhs) is positive.

therefore the right hand side (rhs) must be positive,

so a^2 must be > 2

when a^2 is > 0, both sides are increasing as a increases.

Clearly, lhs > rhs when a^ = 2, and rhs > lhs when a^2 is a lot bigger, so the rhs gets bigger quickly. I expect that they will equal at only one point.

I noticed, while writing this, that the point at which they are equal is a = 2. The lhs = 4^3 = 64. The rhs = 2*2^5 = 64

so the only other solution is a = 2, which means N = +/- sqrt (2).

Verifying.

If N = sqrt(2),

then N^3 + 2N = 4sqrt(2). 5th root = sqrt(2)

N^3 - 2N = 2sqrt(2). 3rd root = sqrt(2)

If N = -sqrt(2),

then N^3 + 2N = -4sqrt(2). 5th root = -sqrt(2)

N^3 - 2N = -2sqrt(2). 3rd root = -sqrt(2)

Only solutions are N = 0 and N = sqrt(2) and N = -sqrt(2)

Interesting enough, in all cases

^{5}√(N^{3} + 2N) = ^{3}√(N^{5} - 2N) = N *Edited on ***April 5, 2009, 1:20 pm**

*Edited on ***April 5, 2009, 1:21 pm**