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 $0$ ?
\pmod{a}
to get $\pmod{a}$. – Arturo Magidin Jul 21 '12 at 04:03