If we have a congruence a≡b (mod m), does it follow that a≡b(mod prime factor of m)

Question

Consider a≡b (mod m) By the divison lemma, a = km+r (0<r<m)

Consider a≡b (mod p, a prime factor of m)

By the division lemma, a = gp + r' (0<r'<p)

m = px (as a prime factor of m)
gp + r' = km + r
((gm)/x) + r' = km + r
m((g/x)-k) = r - r'

Now if the remainders were the same, then (g/x)-k = 0. I'm not sure how to prove this, or where to go from here..

Answer 1

The answer to the title question is yes.

$a\equiv b\pmod m\iff a-b=km$ for some $k\in\mathbb Z.$

Now if $m=pq$ for some $q\in\mathbb Z$, then $a-b=kqp=jp$ for some $j=kq\in\mathbb Z$,

so $a\equiv b\pmod p$.