I have been stuck in a problem related to modular arithmetic. I have tried it using the generalized Euler's formula for $\gcd(a,b)=as+bt$, but have not reached the proof so far.
The question is:
Let $ a, b, n, n' $ all belong to integers($ \mathbb Z $) with $ n > 0 > , n' > 0 $ and $ \gcd(n, n') = 1 $. Show that if $ a \equiv b \mod n $ and $ a \equiv b \mod {n'} $ then $ a \equiv b \mod {nn'} $.
If $a \equiv b \mod n$ and $a \equiv b \mod n^\prime$, then there are integers $k$ and $k^\prime$ such that $kn = a - b = k^\prime n^\prime$. If $kn = k^\prime n^\prime$ and $(n,n^\prime) = 1$, then what is $(n,k^\prime)$?