Find the value of $x^5 + \frac{1}{x^5}$ - question about correctness of method The task is: if  $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.
I used the binomial formula and proved that  $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the the first equality, I get:
$x^2 +2 + \frac{1}{x^2} = 1$ so $x^2 +\frac{1}{x^2} = -1$. Now, the sum of 2 squares is nonnegative and the right side is negative, so when I come to this part does it mean that this method is wrong?
In general, when proving such equalities, when are we allowed to take square (and we don't know if one side of equality is positive as in this task)? Thanks in advance.

$x+\frac{1}{x}=1$ gives $x^2-x+1=0,$ which gives $x^3+1=0$ or $x^3=-1$.
Thus, by your work $$x^5+\frac{1}{x^5}=-x^2-\frac{1}{x^2}=1.$$
Now, about your last question.
You used the following statement.
If $x+\frac{1}{x}=1$ so $\left(x+\frac{1}{x}\right)^2=1$.
It's true because if $a=b$ so $a-b=0$ and from here
$$a^2-b^2=(a-b)(a+b)=0,$$ which gives $$a^2=b^2.$$
Id est, we proved the following claim:
$$a=b\Rightarrow a^2=b^2$$ is true.
