i just computed the solution mod 2 and mod 3 a la chinese remainder theorem the polynomial is =0 mod2 and =0 mod3 so its =0 mod6 n^6 + n^3 + 2n^2 + 2n (mod 2)

i just computed the solution mod 2 and mod 3 a la chinese remainder theorem the polynomial is =0 mod2 and =0 mod3 so its =0 mod6 n^6 + n^3 + 2n^2 + 2n (mod 2) = n^6 + n^3 + 0 + 0 = n^3(n^3+1) = 0*1 or 1*0 = 0 because consecutive numbers are even then odd then even .... for mod3 you can make a table you could also factor the polynomial and see the solution easily n(n+1)(n^2-2n+2)(n^2+n+1)