Prove that $b \equiv c \pmod d$ where $d=\gcd(m,n)$

Question

If $a \equiv b \pmod m $ and $ a \equiv c \pmod n$, Prove that $b \equiv c \pmod d$ where $d=\gcd(m,n)$

We have $a-b=km$ where $k$ is an integer And $a-c=ln$ where $l$ is an integer Now $b-c=ln-km$ What I need to do after this

Answer 1

$$(a-b=km)\wedge (b-c=\ell n)\implies a-c=km+\ell n.$$ Now by definition, the ideal $\:\bigl\{km+\ell n\mid k,\ell\in\mathbf Z\bigr\}\:$ is the ideal generated by $d=\gcd(m,n)$. Thus $a-c\in d\mathbf Z$.

Answer 2

$a\equiv b\pmod m \implies a= b + pm\\ a\equiv c\pmod n \implies a= c + qn$

Setting $a= a$ gives us $b + pm = c + qn$

$b \equiv c \pmod d$ if $d|pm$ and $d|qn$