Problem source

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 source

\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 $r = 1,2,3$