# In congruence classes Z/(mZ), reduce the equation a_m*x_m^2=c_m either by finding convenient representation for a_m and b_m or by using the inverse of

In congruence classes $\frac{Z}{mZ}$, reduce the equation ${a}_{m}\cdot {x}_{m}^{2}={c}_{m}$ either by finding convenient representation for ${a}_{m}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{m}$ or by using the inverse of ${a}_{m}$. Then find a solution for this congruence directly or by replacing ${c}_{m}$ : with its appropriate representative in $\frac{Z}{mZ}$. If there is no solution explain why. Here ${a}_{m},{b}_{m},{x}_{m}\left(=x\right),{c}_{m}\in \frac{Z}{mZ}:$
$In\frac{Z}{19Z},\left[2\right]\cdot {x}^{2}=\left[13\right]:$
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Step 1
Let $y={x}^{2}$ then the given congruence relation becomes,

This congruence has unique solution if and only if gcd(a,m)=1
Here gcd(2,19)=1 hence the given congruence has unique solution.
Now since 2 has an inverse, we get $y\equiv {2}^{-1}13\text{mod}19$ which is the only solution.
The inverse of $2\in {\mathbb{Z}}_{19}$ is 10.
$y\equiv 130\text{mod}19⇒y=16$
Now $y={x}^{2}⇒16={4}^{2}$. Hence x =4 is the required solution.