(=>) first take a≡b(mod m)
so a=b+km , and b=qm+r,0<=r<m.
therefore, a=b+km=(qm+r)+km=(q+k)m+r
hence,a has the same remainder as b.
(<=) let a=q1m+r , b=q2m+r 0<=r<m
a-b=q1m+r-q2m+r=(q1-q2)m
so,m|(a-b) ,we have a≡b(mod m)
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