Square root of 1 I know this is a dumb question but I can't get the answer to another follow up question,


*

*What is the square root of 1?

*If the square root of 1 is itself then why does other square root of number not equal to themselves?

*Is there other square root of 1 aside from 1?


This question is somewhat a deconstruction of the fundamental logic of mathematical concepts but I can't find a good reason to answer this questions. Hoping for a good reason.
[Edit] if we're going to talk about principal square roots then therefore -1 would be included but i'm only referring to non negative numbers therefore 
3.1. if there are other square root of 1 which could be between 0 and 1?
 A: *

*The square root of $1$ is $1$.

*This is because the number $1$ has the special property $1\cdot 1=1$. This is not true for $2$, for instance: $2\cdot 2$ is not $2$.

*Yes, there is another number $x$ with $x^2=1$, namely $x=-1$. Usually, this solution is not called a square root of $1$, however, because you want "square root" to be a function, so that its value has to be unique.
Edit: To see that there are no further solution to $x^2=1$, notice that $x^2=1$ implies $(x+1)(x-1)=0$ and thus $x=\pm 1$.
A: In a field such as $\,\mathbb Q,\ \mathbb R,\ \mathbb C,\,$ we have $ \ x^2 = 1 \iff (x-1) (x+1) = 0\iff x = \pm 1.\, $ In rings that are not fields there can be more than two square-roots, e.g. modulo $15$ there are two additional roots $ \ (\pm\,4)^2\equiv 1\pmod{\!15}.\,$ In some contexts authors define single-valued square-root functions that uniformly select one of the roots, e.g. the non-negative root, or principal branch, etc.
In ring theory we can adjoin a "sign" = generic square root of one $\,s = \pm1\,$ to a ring $ \,R\,$ as follows $ R[s]\cong R[x]/(x^2\!-\!1).\,$ If $\,2\,$ is invertible, CRT $\Rightarrow R[s] \cong R[x]/(x\!-\!1) \times R[x]/(x\!+\!1)\ \cong R^2,\, $ via $\, f(x)\to (f(1),f(-1));\,$ informally, this ring product implements arithmetic in two parallel universes (rings), where in the first world we have $\,s = 1\,$ , and $\, s = -1\,$ in the second.
As to your second question concerning why there aren't other numbers that are their own square root, note that $ \ \sqrt{x} = x\ \Rightarrow x = x^2\ \Rightarrow\ x\ (x-1) = 0\ $ so $ \,x = 0\,$ or $ \,x = 1\,$ in a field (or domain). Elements satisfying $ \ x^2 = x\ $ are known as idempotents. Idempotents that are nontrivial ($\ne\, 0,1$) are intimately connected with nontrivial factorizations. In the above example $\,\mathbb Z/15 =$ integers $\!\bmod 15,\,$ we have a factorization $ \mathbb Z/15 \cong \mathbb Z/3 \times \mathbb Z/5\,$ arising from idempotent $\,6,\,$ i.e. $ (0,1)\in \mathbb Z/3 \times \mathbb Z/5.\,$ Note that if $ \,e\,$ is idempotent then $ \,(1\!-\!2\,e)^2 = 4\,(e^2\!-\!e)+1 = 1,\, $ so $ \,2e\!-\!1\,$ is a square-root of $1;\,$ it is a nontrivial square root $ \,2e\!-\!1\ne \pm 1\,$ if $\,2\,$ is cancellable. In our example $ \,e=6,\,$ so $ \,1\!-\!2e \equiv 4\,$ which does indeed square to $ \,1\pmod{\!15}.\,$ We can quickly factor $15$ from this nontrivial sqrt $ \,s,\,$ namely $ \,\gcd(15,s-1) = \gcd(15,3) = 3\,;$ $ \ \gcd(15,s+1) = \gcd(15,5) = 5.\,$ In fact many integer factorization algorithms work by searching for nontrivial square-roots. This way of factoring by simple gcd computations works more generally. One can quickly factor $ \,n\,$ given any polynomial with more roots $\!\bmod n\,$ than the degree of the polynomial - see my post here.
A: *

*The square root of $1$ is $\pm1$.

*$1$ is the only number which when multiplied by itself equals $1$.

*$-1$ multiplied by itself equals $1$.
