Reducing yn+(1+y)n+(1−y)n=0y^n+(1+y)^n+(1-y)^n=0yn+(1+y)n+(1−y)n=0 modulo 222 and using odd nnn gives y+(1+y)+(1−y)=y+2≡0(mod2)y+(1+y)+(1-y)=y+2\equiv0\pmod2y+(1+y)+(1−y)=y+2≡0(mod2). Thus yyy is even.