Exercise 6.4.6

Proof. Let $f=x^5-6x+3$.

(a)

Let $\sigma \in S_5$ a permutation of order 5. Write $\sigma ={\sigma }_1\cdots {\sigma }_r({\sigma }_i\ne e)$ the cycle decomposition of $\sigma$. Let $d_i=|{\sigma }_i|$ the order of ${\sigma }_i$ in $S_n$. As the cycles are disjoint, for all integer $k$, ${\sigma }^k={\sigma }_1^k\cdots {\sigma }_r^k$ and $$\begin{array}{llll}\hfill {\sigma }^k=e& ⟺{\sigma }_1^k=\cdots ={\sigma }_r^k=e& \hfill & \\ \hfill & ⟺d_1\mid k,d_2\mid k,\dots ,d_r\mid k& \hfill & \\ \hfill & ⟺\mathrm{lcm}(d_1,\dots ,d_r)\mid k& \hfill & \end{array}$$

So the order of $\sigma$ is the lcm of the orders $d_i$.

$$5=\mathrm{lcm}(d_1,\dots ,d_r).$$

As $d_i\mid 5,i=1,\dots ,r$, and $d_i\ne 1$, where $5$ is prime, $d_i=5$. The cycles ${\sigma }_i$ being disjoint, as $d_i=|{\sigma }_i|=\mathrm{length}({\sigma }_i)$, $d_1+\cdots +d_r\le 5$, thus $r{d}_1=5r\le 5$, so $r=1$.

Conclusion: $\sigma ={\sigma }_1$ is a 5-cycle.

(b)

Let $f:ℝ\to ℝ,x\mapsto f(x)=x^5-6x+3$.

If $x\in ℝ$, $f^{\prime }(x)=5x^4-6<0⟺x^4<\frac{6}{5}⟺-x_0<x<x_0,\mathrm{where}{x}_0=\sqrt[4]{\frac{6}{5}}$.

$f$ is so strictly increasing on $]-\infty ,-x_0]$, strictly decreasing on $[-x_0,x_0]$, and strictly increasing on $[x_0,+\infty [$.

$f(x_0)=x_0(x_0^4-6)+3=x_0(\frac{6}{5}-6)+3=-\frac{24}{5}{x}_0+3=\frac{3}{5}(5-8x_0)<0$: indeed $x_0=\sqrt[4]{\frac{6}{5}}>1>\frac{5}{8}$, so $5-8x_0<0$.

$f(-x_0)=-x_0(x_0^4-6)+3=\frac{24}{5}{x}_0+3>0$.

As $f$ is continuous, $\underset{x\to -\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x)=-\infty ,f(-x_0)>0$, and $f$ is strictly increasing on $]-\infty ,-x_0]$, the Intermediate Values Theorem shows that $f$ has a unique root in $]-\infty ,-x_0]$.

With a similar reasoning on $[-x_0,x_0]$ and on $[x_0,+\infty [$, with $f(-x_0)>0,f(x_0)<0,\underset{x\to +\infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x)=+\infty$, we prove that $f$ has a unique root in $[-x_0,x_0]$, and also in $[x_0,+\infty [$.

Conclusion: $f$ has exactly three real roots.