Is it true that "there is no such thing as the square root of minus one"? Is the statement "there is no such thing as the square root of minus one" a true statement?
It seems to me that we need to be careful about the word "the" as it appears in the statement. If we see it as implying uniqueness, then surely the statement is true after all, since $i$ and $-i$ are distinct square roots of minus one.
 A: Indeed, as suggested in the question, if we were to speak of the root of $-1$, then there should be just one such value, or at least one of the two possible values should be distinguished.
For positive reals, we can speak of, for instance, the square root of $4$, because out of the two possible values of $+2$ and $-2$, only one is positive (and this is the one we take). Note that numbers $+2$ and $-2$ are quite different: for instance, $+2$ is a square of a real number, and $-2$ is not. 
For $-1$, things are a little trickier. There are two possible choices of the root, $+i$ or $-i$. And in some sense, these choices are both equally legitimate. There is no way to distinguish $+i$ from $-i$ using only arithmetic operations and knowing what the real numbers are. This is a consequence of the fact that the conjugation mat $z \mapsto \bar{z}$ preserves multiplication, addition, and leaves $\mathbb{R}$ invariant.
A: Usually we speak of 'the' square root of a negative number $-k$ ($k\in\mathbb R_{>0}$) as $\sqrt{-k}=\sqrt{k}\cdot i$, but we could have chosen $-\sqrt k\cdot i$ just as well. This follows from the fact that the Galois Group of $\mathbb C$ contains two elements (two automorphisms of $\mathbb C$), $e$ and $s$:
$$
e:\mathbb C\to \mathbb C: c\mapsto c\\
s:\mathbb C\to \mathbb C: c\mapsto \overline {c}
$$
This means that $\mathbb C$ would be exactly the same as (undistinguishable from) $s(\mathbb C)$, where $i$ and $-i$ are interchanged. 
A: Presumably you do not have an issue with the existence of $i$ and $-i$, and are just debating the grammar of the sentence. Note that the square root function is, in general, defined to be the positive root or, in such cases that sign does not make sense, we use the principal solution. That is, the value of $e^{\pi/2+k\pi}$ with smallest positive argument (namely, $i$).  
Using this convention, there is such a thing as the square root of $-1$. But as you note, there are $2$ numbers which square to $-1$.  
A: It depends on how you define "squareroot"...
But consider rotations around the $z$-axis over an angle $\phi$.
We may denote such rotations as $R(\phi)$,
and we have the property $R^2(\phi) = R(2\phi)$.
Now consider $R(180^\circ) = R(-180^\circ)$ - that would be $-1$, so $\textbf{i} = R^2(\pm 90^\circ)$ has the property that $\textbf{i}^2 = -1$.
Or $\sqrt{R(\pm 180^\circ)} = R(\pm 90^\circ)$.
Or $\sqrt{-1} = \pm \textbf{i}$.

Note that the squareroot is double-values as
$$x^2 = z \Rightarrow x = \pm \sqrt{z}$$
