Suppose $x$ is a perfect square mod $n$. If $xy \equiv 1 $ mod $n$, prove $y$ is a perfect square also. Here's what I have:
If $a$ is a perfect square mod $n$, then $\exists m \in \mathbb{Z}^+$ such that $m^2 \equiv a \text{ mod } n$. We want to find some $k$ such that $k^2 \equiv b \text{ mod } n$.
\begin{align*}
ab &\equiv 1 \text{ mod } n \\
m^2b &\equiv 1 \text{ mod } n \\
b &\equiv  m^{-2} \text{ mod } n
\end{align*}
Is this satisfactory? Thanks!
 A: We have $a \equiv m^2 \mod n$.  And we have $ab \equiv 1 \mod n$.
So $(mb)^2 \equiv m^2b^2 \equiv ab^2 \equiv (ab)b \equiv b \mod n$.
A: This is not an improvement on fleablood's Answer but might give you some insights.... All values here are integers.
Theorem. For all non-zero $x,y$:  There exist $u,v$ such that $ux+vy=1$ iff $\gcd(x,y)=1.$ Equivalently, there exists $u$ such that $ux\equiv 1\pmod y$ iff $\gcd(x,y)=1.$  
Corollary:  The Fundamental Theorem of Arithmetic (FTA) : For all non-zero $x,y,z,$ if $y|\;xz$ and $\gcd(y,x)=1$ then $y|z.$ Because,  with $ux+vy=1$ we have $y|\;u(xz)=(ux)z=(1-vy)z=z-y(vz)\implies y|z.$ 
By the Theorem,  $ab\equiv 1\pmod n\implies \gcd(a,n)=1.$Now since $a-m^2\equiv 0\pmod n$ for some $m,$ we have $\gcd(m,n)=1.$
.... Because if $p|\;m$ and  $p|\;n$ then $p|\;m^2 ,$ hence  $p|\; (a-m^2) \land p|\;m^2,$ so $p|(a-m^2)+m^2=a$... But $(p|a\land p|\;n)\implies p=\pm 1 .$
Since $\gcd(m,n)=1,$ the  Theorem  implies  there exists $u$ with $um\equiv 1 \pmod n.$  Now $1\equiv u^2m^2\equiv u^2a\equiv ba \pmod n.$ So $n|\;(u^2-b)a.$ 
Since $\gcd(a,n)=1,$ the FTA implies $n|(u^2-b).$   
