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mathisfun1 · #2 July 18th, 2007, 12:40 PM

1) We'll show that f(1)=0; the result follows. Write $f(x^5) = (x^4+x^3+x^2+x+1)r(x)-xg(x^5)-x^2h(x^5) \ \ (*)$ . Note that the RHS can only have powers of x that are multiples of 5. $xg(x^5)$ and $x^2h(x^5)$ 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 $r(x)=a_0+a_1x+a_2x^2+...+a_nx^n$ . The first term of the RHS of (1) is $a_0$ . Write the highest power of x that is a mulitple of 5 as $5k$ . Consider when k=1. Now, the $x^4$ term has coefficient $\Sigma_{i=0}^m a_i=0$ . If m<4 then the coefficient for $x^5$ is $\Sigma_{i=1}^m a_i = -a_0$ , in which case x=1 is a root for f(x). If m=4 then there are there are two possibilities. If $x^5$ has coefficient $\Sigma_{i=1}^4 a_i = -a_0$ then we're done. If instead the coefficient is $\Sigma_{i=1}^{5} a_i = a_5-a_0$ , then $f(1)$ (the sum of the coefficients of $x^0$ and $x^5$ ) equals $a_5$ , in which case we're done since $a_5=0$ (being the coefficient of $x^9$ . Induct on k.