Prove that ## b\equiv c\pmod {n} ##, where the integer is....

[Math100]

Homework Statement

If $a\equiv b\pmod {n_{1}}$ and $a\equiv c\pmod {n_{2}}$, prove that $b\equiv c\pmod {n}$, where the integer $n=gcd(n_{1}. n_{2})$.

Relevant Equations

None.

Proof:

Suppose $a\equiv b\pmod {n_{1}}$ and $a\equiv c\pmod {n_{2}}$ where the integer $n=gcd(n_{1}, n_{2})$.
Then $a-b=n_{1}k_{1}$ and $a-c=n_{2}k_{2}$ for some $k_{1}, k_{2}\in\mathbb{Z}$.
This means $b-c=n_{2}k_{2}-n_{1}k_{1}$.
Since $n=gcd(n_{1}, n_{2})$, it follows that $n_{1}=qn$ and $n_{2}=tn$ with $gcd(q, t)=1$.
Thus $b-c=tnk_{2}-qnk_{1}=n(tk_{2}-qk_{1})\implies n\mid (b-c)\implies b\equiv c\pmod {n}$.
Therefore, $b\equiv c\pmod {n}$ if $a\equiv b\pmod {n_{1}}$ and $a\equiv c\pmod {n_{2}}$, where the integer $n=gcd(n_{1}, n_{2})$.

 

[fresh_42]

That's fine.

Maybe a bit more intuitive is the following order of arguments (same as yours):

$n\,|\,n_1 \,|\,(a-b)$ so $a\equiv b\pmod n$ and for symmetry reasons $a\equiv c\pmod n.$ Now $\equiv$ is an equivalence relation (reflexive, symmetric, transitive), hence $b\equiv c \pmod n$ which is the transitivity property.