Finding at least 2 elements in a set that satisfys an equation For any integers $n>2$, show that there are at least 2 elements in $U \big(n\big)$  that satisfy $x^{2}=1$
Looking for a vital hint to this question.
My intuition tells me that I ought to try for small n values and ascertain that there are in fact at least 2 elements for any small n values I choose.
Since U(n) is by definition the set of positive integers less than n and relatively prime to n, any n values greater than the small n values for which I have checked to yield at least 2 elements that satisfy $x^{2}=1$ must hold for all any n(large) values
The element x=1 will indeed satisfy $x^{2}=1$.
 A: Let $n\ge 2$.  The group $U(n)$ of units modulo $n$ can be thought of as having as elemwnts the positive integers less than $n$. The group operation is multiplication modulo $n$. So the product $ab$ of two elements $a$ and $b$ of $U(n)$ is the remainder when the ordinary product is divided by $n$.
Now we turn to the solving the equation $x^2=1$ in $U(n)$.  There is the obvious solution $1$, since the remainder when the ordinary product $(1)(1)$ is divided by $n$ is $1$. Or, to use "mod" language, we have $1^2\equiv 1\pmod{n}$.
For all $n\ge 2$, $n-1$ is a solution of $x^2=1$. To show this, we show that the remainder when the ordinary product $(n-1)^2$ is divided by $n$ is $1$. Note that $(n-1)^2=n^2-2n+1=(n-2)(n)+1$. So $(n-1)^2$ is $1$ more than a multiple of $n$. In "mod" language we have $(n-1)^2\equiv \pmod{n}$.
If $n\gt 2$, then $n-1\ne 1$, so we have exhibited two different solution of $x^2=1$ in $U(n)$. For completeness we should perhaps explicitly verify that $n-1$ is an element of $U(n)$. 
Remark: It turns out that if $n$ is a prime $\gt 2$, or a power of an odd prime, then there are exactly two solutions of $x^2=1$ in $U(n)$.  
But if $n$ is the product of $k$ distinct odd primes, then the equation $x^2=1$ has $2^k$ solutions in $U(n)$. Or, as I would prefer to put it, the congruence $x^2\equiv 1\pmod{n}$ has $2^k$ solutions.  As a small illustration, it is not hard to check that $x^2=1$ has the solutions $1$, $4$, $11$, and $14$ in $U(15)$. 
