I have an ugly way to solve this problem.But I believe it is not the most beutiful and simple one.I am busy these days,Maybe I can post the complete solution some days.But now only an hint: Notice that the degree of LHS and RHS is 5 and 4,so we may use some inequality: Case 1.$x<0,y<0$,then $LHS<0,RHS>0$ Case 2.$x\ge 3,y\ge3$,$LHS\ge RHS$ Case 3.$x>0,y<0$.first we get $x+y\le-1$,and we can deduce that there must exist $A$ such that when $x\ge A$ $LHS<RHS$( I didn't compute carefully,maybe$A=4$ can work. The other case is easy.(we only need to solve a few equation with one variable.) I am sorry for the not complete solution

Hawk Tiger wrote: I have an ugly way to solve this problem.But I believe it is not the most beutiful and simple one.I am busy these days,Maybe I can post the complete solution some days.But now only an hint: Notice that the degree of LHS and RHS is 5 and 4,so we may use some inequality: Case 1.$ x < 0,y < 0$,then $ LHS < 0,RHS > 0$ Case 2.$ x\ge 3,y\ge3$,$ LHS\ge RHS$ Case 3.$ x > 0,y < 0$.first we get $ x + y\le - 1$,and we can deduce that there must exist $ A$ such that when $ x\ge A$ $ LHS < RHS$( I didn't compute carefully,maybe$ A = 4$ can work. The other case is easy.(we only need to solve a few equation with one variable.) I am sorry for the not complete solution Your solution is not full. Try again with result: If $ x+y+z=0$ then $ 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ Imply that it contain term $ x+y+1$ .

Solution was posted here: http://www.mathlinks.ro/viewtopic.php?t=166171

$ 5|x^5 + y^5 +1 $ which is impossible because $x^5$ mod 5 can be just 1 or 0