If a divides $b-1$ and a divides $c-1$ then a divides $bc-1$

Question

I am wondering if the proof I did for this problem is correct. We know a divides b-1

so $b-1=a(t)$ for some integer t

also a divides c-1

so $c-1=a(r)$ for some integer r, so by this logic

so then $b=a(t)+1$ and $c=ar+1$

So then $b(c)-1=(at+1)(ar+1)-1$

$bc-1=a^2tr+at+ar+1-1$

so then subtract the 1

$bc-1=(a^2tr+at+ar)=a(atr+t+r)$

so $a$ is a factor of $bc-1$ proof complete.

Answer 1

$$ (b-1)(c-1) +(b-1) + (c-1) \; \; = \; \; bc-1 $$