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 a_m. Then find a solution for this congruence directly or by replacing c_m : with its appropriate representative in Z/(mZ). If there is no solution explain why. Here a_m, b_m, x_m(=x),c_m in Z/(mZ): In Z/(19Z), [2]*x^2=[13]:

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 a_m. Then find a solution for this congruence directly or by replacing c_m : with its appropriate representative in Z/(mZ). If there is no solution explain why. Here a_m, b_m, x_m(=x),c_m in Z/(mZ): In Z/(19Z), [2]*x^2=[13]:

Question
Congruence
asked 2021-01-30
In congruence classes \(\displaystyle\frac{{Z}}{{{m}{Z}}}\), reduce the equation \(\displaystyle{a}_{{m}}\cdot{{x}_{{m}}^{{2}}}={c}_{{m}}\) either by finding convenient representation for \(\displaystyle{a}_{{m}}{\quad\text{and}\quad}{b}_{{m}}\) or by using the inverse of \(\displaystyle{a}_{{m}}\). Then find a solution for this congruence directly or by replacing \(\displaystyle{c}_{{m}}\) : with its appropriate representative in \(\displaystyle\frac{{Z}}{{{m}{Z}}}\). If there is no solution explain why. Here \(\displaystyle{a}_{{m}},{b}_{{m}},{x}_{{m}}{\left(={x}\right)},{c}_{{m}}\in\frac{{Z}}{{{m}{Z}}}:\)
\(\displaystyle{I}{n}\frac{{Z}}{{{19}{Z}}},{\left[{2}\right]}\cdot{x}^{{2}}={\left[{13}\right]}:\)

Answers (1)

2021-01-31
Step 1
Let \(\displaystyle{y}={x}^{{2}}\) then the given congruence relation becomes,
\(\displaystyle{a}{y}\equiv{b}\text{mod}{m}{t}\hat{{i}}{s}{2}{y}\equiv{13}\text{mod}{19}\)
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 \(\displaystyle{y}\equiv{2}^{{−{1}}}{13}\text{mod}{19}\) which is the only solution.
The inverse of \(\displaystyle{2}\in\mathbb{Z}_{{{19}}}\) is 10.
\(\displaystyle{y}\equiv{130}\text{mod}{19}\Rightarrow{y}={16}\)
Now \(\displaystyle{y}={x}^{{2}}\Rightarrow{16}={4}^{{2}}\). Hence x =4 is the required solution.
0

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