1 problem found
Let \(\mathrm{p}_{0}(x)=(1-x)(1-x^{2})(1-x^{4}).\) Show that \((1-x)^{3}\) is a factor of \(\mathrm{p}_{0}(x).\) If \(\mathrm{p}_{1}(x)=x\mathrm{p}_{0}'(x)\) show, by considering factors of the polynomials involved, that \(\mathrm{p}_{0}'(1)=0\) and \(\mathrm{p}_{1}'(1)=0.\) By writing \(\mathrm{p}_{0}(x)\) in the form \[ \mathrm{p}_{0}(x)=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}+c_{6}x^{6}+c_{7}x^{7}, \] deduce that \begin{alignat*}{2} 1+2+4+7 & \quad=\quad & & 3+5+6\\ 1^{2}+2^{2}+4^{2}+7^{2} & \quad=\quad & & 3^{2}+5^{2}+6^{2}. \end{alignat*} Show that we can write the integers \(1,2,\ldots,15\) in some order as \(a_{1},a_{2},\ldots,a_{15}\) in such a way that \[ a_{1}^{r}+a_{2}^{r}+\cdots+a_{8}^{r}=a_{9}^{r}+a_{10}^{r}+\cdots+a_{15}^{r} \] for \(r=1,2,3.\)
Solution: \begin{align*} && p_0(x) &= (1-x)(1-x^2)(1-x^4) \\ &&&= (1-x)(1-x)(1+x)(1-x^2)(1+x^2) \\ &&&= (1-x)^2 (1+x)(1-x)(1+x)(1+x^2) \\ &&&= (1-x)^3 (1+x)^2 (1+x^2) \end{align*} \begin{align*} && p_0'(x) &= 3(1-x)^2(1+x)^2(1+x^2) + (1-x)^3 q(x) \\ \Rightarrow && p_0'(1) &= 3 \cdot 0 \cdots + 0 \cdots \\ &&&= 0 \\ && p_1'(x) &= p_0(x) + xp'_0(x) \\ \Rightarrow && p_1'(1) &= p_0(1) + 1\cdot p_0'(1) \\ &&&= 0 + 1 \cdot 0 \\ &&&= 0 \end{align*} Notice that \(p_0(x) = (1-x-x^2+x^3)(1-x^4) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7\), so: \(p'_0(x) = -1-2x+3x^2-4x^3+5x^4+6x^5-7x^6 \Rightarrow p'_0(1) = 0 = -1 -2 -4 -7 + 3 + 5+6\). \((xp'_1(1))' = 0 = -1^2-2^2-4^2-7^2 + 3^2 + 5^2 + 6^2\). Consider \(q_0(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\), then \((1-x)^4\) is a factor, so in particular we know \(q_0(1), (xq_0(x))'|_{x=1} = 0,(x(xq_0(x))')'|_{x=1} = 0\), and so: \(q_0(x) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7 - x^8+x^9+x^{10}-x^{11}+x^{12}-x^{13}-x^{14}+x^{15}\), and so: \(1^r+2^r+4^r+7^r+8^r+11^r+13^r+14^r = 3^r+5^r+6^r+9^r+10^r+12^r+15^r\) for \(r = 1,2,3\)