<s>[INST] More than two solutions to $x^2 -1 = 0$ in $\Bbb{Z}/n\Bbb{Z}$ if $n$ is odd and not a power of a prime 
Let $n$ be a positive odd integer such that it is not the power of a single prime, show that in  $\Bbb{Z}/n\Bbb{Z}$ there are more than two solutions to $x^2 -1 = 0$

The requirement that $n$ is not the power of a single prime is stumping me.  Does that mean that we can write n in the form of $p\times q^s\times\dots$ ?
 [/INST] Let $n=ab$ with $\gcd(a,b)=1$ and $a,b>1$ each odd, By the Chinese remainder theorem we can choose $x$ in $\{1,...,n\}$ which is $-1$ mod $a$ and is $+1$ mod $b$. Since $a,b$ are odd, this $x$ will not be $\pm 1$ mod $n.$ 
However we have $x^2=1$ both mod $a$ and mod $b$, hence mod $n=ab.$ [ This uses that if $\gcd(a,b)=1$ and each of $a,b$ divides some $n$ then also $ab$ divides $n.$ ] So we have three solutions $-1,1,x$ to $x^2-1=0$
A detail: Since $x=-1$ mod $a,$ $a$ is a divisor of $x+1.$ Then using that $x+1$ is a divisor of $x^2-1,$ we have that $a$ divides $x^2-1$ since if $u$ divides $v$ and $v$ divides $w$ then $u$ divides $w.$ Similarly from $x=1$ mod $b$ we have $b$ divides $x-1$ and then since $x-1$ divides $x^2-1,$ we have $b$ divides $x^2-1.$ So at this point we have $\gcd(a,b)=1$ and each of $a,b$ divides $x^2-1,$ and it is from this that we can conclude $ab$ divides $x^2-1.$ [I didn't bring out before that $a,b$ each divide the same thing.] Finally to say $ab$ divides $x^2-1$ is the same, since $n=ab,$ as saying $x^2=1$ mod $n.$ I think it is clear that $x$ here is neither of $-1,1$ mod $n$ so we indeed have at least three solutions to $x^2=1$ mod $n.$
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