1) We'll show that f(1)=0; the result follows. Write . Note that the RHS can only have powers of x that are multiples of 5. and have only powers of x congruent to 2 and 3 mod 5, respectively. Choose coefficients for the RHS of (*) such that the terms with powers of x congruent to 3 and 4 (mod 5) vanish. Write . The first term of the RHS of (1) is . Write the highest power of x that is a mulitple of 5 as . Consider when k=1. Now, the term has coefficient . If m<4 then the coefficient for is , in which case x=1 is a root for f(x). If m=4 then there are there are two possibilities. If has coefficient then we're done. If instead the coefficient is , then (the sum of the coefficients of and ) equals , in which case we're done since (being the coefficient of . Induct on k.