The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$.  The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain complex values
Yet the reasoning for the implication itself is
\begin{align}
x+\frac {1}{x}&=1 \\
\color {green}{x^2}&=\color {green}{x-1 }\\
x^3&=\color {green}{x^2}-x \\
&=(\color {green}{x-1})-x \\
&=-1.
\end{align}
Hence $x^6=1$, which implies $x^7=x$.
I had just started reading this book when it was presented, asking you to prove the implication without finding the roots first.
So my question: is this implication really justified?   Should the author have said to only prove it for an implicit (complex) specific value of $x$?  If I didn't know any better, I'd say he presented it like it was true for some range of values...he couldn't of...
 A: Define two propositions: $p:= x+\frac{1}{x}−1=0$ and $q:=x^7+\frac{1}{x^7}−1=0$. You are trying to find the values of $x$ for which $p \wedge q$ is true. The actual question is 'prove that $p \Rightarrow q$ is true'. These two propositions are different.
If you try to solve $x+\frac{1}{x}=1$ you end up with $x = e^{i\frac{\pi}{3}}$ and $x = e^{-i\frac{\pi}{3}}$. Now, in both cases
$$x^7+\frac{1}{x^7} = 2\cos \frac{7\pi}{3} = 2\cos \frac{\pi}{3} = 1.$$
A: Let $x_n:=x^n+\frac{1}{x^n}$ so $x_{n+1}=x_1 x_n - x_{n-1}=x_n - x_{n-1}$ and $x_{n+2}=x_{n+1}-x_n=-x_{n-1}$. Thus $x_7=(-1)^2x_1=1$.
A: The given equation implies $x^2-x+1=0$; multiplying by $x+1$, we get $x^3+1=0$. This implies $x^6=1$ and therefore
$$
x^7=x
$$
Your argument is sound as well.
The key is that a solution to $x+\frac{1}{x}=1$ is a 6th root of $1$. Indeed,
$$
x^6-1=(x^3-1)(x^3+1)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
$$
Thus any root of $x^2-x+1$ is also a root of $x^6-1$.
A: You shouldn't be trying to find where $y_1=y_2$, but whether a root of $y_1$ is necessarily a root of $y_2$.
That is,
$$\forall x\Big(y_1(x)=0\implies y_2(x)=0\Big)$$
or equivalently,
$$\forall x\Big(y_1(x)\neq0\lor y_2(x)=0\Big)$$
