Question

Question:- If (a,n)= 1and s,t are solution of $\mathbf{\text{ax}}\mathbf{\equiv }\mathbf{\text{b(modn)}}$ prove that$\mathbf{\text{s}}\mathbf{\equiv }\mathit{t}\mathbf{\text{(modn)}}$ [Hint show that t and use theorem ]

Short answer

Answer:-

In the below solution, it has been proved that the $s\equiv t(mod n)$.

Step-by-step solution

Conceptual Introduction

Let $m_1, m_2 ....m_f$be pair wise relatively prime positive integers, (it means that $\left(m_i,m_j\right)=1$, whenever $i\ne j$) . Assume that the$a_1, a_2....., a_r$ , are any integers.

Then the system,

$$\begin{gathered} x\equiv a_1\left(mod m_1\right) \\ x\equiv a_2\left(mod{m}_2\right) \\ x\equiv a_3\left(mod{m}_3\right) \\ . \\ . \\ . \\ x\equiv a_r\left(mod{m}_r\right) \end{gathered}$$

has a solution.

By using the theorem 1.4, and as  (a,n ) =1 the algebraic equation of the congruent is

The congruent equation is:

$x\equiv b(mod n)$

It can be rewritten as an algebraic equation, with congruent replaced by an equal to sign.

$$\begin{gathered} x\equiv b(mod n) \\ ax-ny=b \end{gathered}$$

Replace the values into the equation

If the value is:

(a,n ) = d

Then, $d=gcd\left(a,n\right)$

Replace s as b and t as $ax-ny=b$, in this equation .

$$\begin{gathered} ax-ny=b \\ sx-ny=t \end{gathered}$$

The algebraic equation is represented as an congruence equation as

$sx\equiv t\left(modn\right)$

By multiplying both sides of the equation by

$asx\equiv ta(mod n)$

Rearrange the terms

$$\begin{gathered} asx\equiv ta(mod n) \\ x=n|\left(as-at\right) \end{gathered}$$

Hence, proved $x=n|\left(as-at\right)$.