Prove that if b1,b2∈Z and d1,d2∈Z+, then there exists at least one solution x∈Z satisfying simultaneously:
x≡b1(mod d1)
x≡b2(mod d2)
if and only if gcd(d1,d2)|(b1−b2).
So, so far what I have done is the following:
x≡b1(mod d1)⇒x=b1+k1d1, some k1∈Z
x≡b2(mod d2)⇒x=b2+k2d2, some k2∈Z
Rearrange to get: (b1−b2)+k1d1=k2d2
But now I'm stuck.
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