If $A \equiv 1 \pmod y$ and $A \equiv 1 \pmod {by}$ where $\gcd(b,y)=1$, then $A \equiv 1 \pmod b$?

Question

I am looking at examples where I am coming across the following situation again and again:

$A \equiv 1 \pmod y$, $A \equiv 1 \pmod {by}$, where $\gcd(b,y)=1$. Then I am always getting $A \equiv 1 \pmod b$.

But I am unable to prove this. Is there a proof of such a result?

Answer 1

If $ A \equiv 1 \pmod {by}$ then $\exists k : \; A = b\, y\,k+1$ so $$A \equiv b \, y \, k + 1 \equiv 0 \, y \, k + 1\equiv 1 \pmod b.$$ You can do the same with $y$ instead of $b$.