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I have never really directly dealt with congruences until I was introduced to the Chinese Remainder Theorem. Although there are tons of different versions of this theorem out there, currently I am more interested in solving congruences. I had some questions regarding steps that were used in solving the following congruent linear equation for $X_1$.

$99X_1\equiv1\pmod8$

Subtracting 96(which is a factor of 8 from both side)

$3X_1\equiv1\pmod 8$ (Subtracting any factor of 8 on R.H.S is like subtracting $0$. So R.H.S remains unchanged)

Adding 8 gives:

$3X_1\equiv(1+8) \pmod8$

$3X_1\equiv9\pmod8$ so $X_1=3$

Now when 8 was added why was it only added to the R.H.S . I thought any factor of 8 whether being subtracted (or added?) on R.H.S is like subtracting or adding $0$ ?

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Use \pmod{a} to get $\pmod{a}$. –  Arturo Magidin Jul 21 '12 at 4:03

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up vote 3 down vote accepted

Precisely because adding $8$ modulo $8$ is the same as adding $0$, you can add $8$ to only one side and not change the congruence (just like you can add $0$ to only one side of $x=y$ to get $x=y+0$ and not change the equality).

Alternatively, since $0\equiv 8\pmod{8}$, you are adding the congruence $$3x_1\equiv 1\pmod{8}$$ to the congruence $$0\equiv 8\pmod{8}$$ to get $$3x_1+0 \equiv 1+8\pmod{8}.$$ This is like adding $x=a$ to $y=b$ in order to get $x+y=a+b$.

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Oh okay , so we can add and subtract factors of 8 but from only one side at a time –  Rajeshwar Jul 21 '12 at 4:16
1  
@Rajeshwar: No, you can add multiples of $8$ to either side, or to both sides, at will, without disturbing the congruence. –  Arturo Magidin Jul 21 '12 at 4:17
    
Thanks for clearing that up –  Rajeshwar Jul 21 '12 at 4:18

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