Assume $b$ is a square root of an element $n$, and let $a$ be an unknown element which is also a square root of $n$.
If $n$ has only one root, then $a = b$.
$a^2 - b^2 = n - n = 0$ so $(a + b)(a - b) = 0$
$a = b$ or $a = -b$
hence $a$ can take the value of $b$ and $-b$, however this contradicts our initial assumption of $n$ having one root, so it has at least two roots.
Since the field is ordered, we have $b ≠ -b$ so they are two different square roots of $n$. From the axiom of an ordered field, we have $b > 0$, $b < 0$ or $b = 0$. Now $b ≠ 0$ as if $b = 0$, then $b^2 = n = 0$, which is against our assumption that $n ≠ 0$. Hence $b > 0$ or $b < 0$, so $n$ has one positive and negative square root, which happen to be additive inverses of each other.
Assume on the other hand that n has three square roots, $a$, $b$ and $c$, where $a$ and $c$ are unknown elements, then
$a^2 = b^2 = c^2 = n$
$(a - b)(a + b) = 0$, $(b - c)(b + c) = 0$, $(a - c)(a + c) = 0$
so $a = b$ or $a = -b$, $b = c$ or $b = -c$, $a = c$ or $a = -c$
Now if $a = c$, we get
$c = b$ or $c = -b$
The same holds for $a = -c$
So we are left with exactly two square roots and not three, by which this argument can be extended if we assume any number of square roots above two.
Therefore if a non-zero element of an ordered field has a square root, then it must have two square roots where one is positive and one is negative.
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