LET L=(x3+20x)1/5 and
Clearly L^5=(x3+20x) and
R=L=x is a solution (maybe not the only one)
SINCE 20*X=X^5-X^3 and x is not a zero
and x^2=5 is the answer.
X1= +sqrt(5); X2=- sqrt(5); real roots
X3= sqrt(-4)=2i; X4=- sqrt(-4)=-2i; complex.
Proving absence of other non-zero roots is elementary (polynom of 5th degree!).
Nice puzzle - rated it 4.
Edited on February 24, 2014, 6:17 am