Does $\,n\mid a_1b_1,\,a_2b_2, a_1b_2+a_2b_1\Rightarrow n\mid a_1b_2$? Imagine that we have two pairs of integers $(a_1,b_1)$ and $(a_2, b_2)$ where
$$ a_1b_1\equiv 0,\,\ a_2b_2\equiv 0,\,\ a_1b_2+a_2b_1\equiv 0\pmod n$$
Does this imply that
$$ a_1 b_2 \equiv 0\pmod n $$
I assume that $a_1 b_1 \equiv 0\pmod n$ is only true if $a_1$ and $b_1$ are divisors of $n$. Using that, I've checked $a_1b_2 = 0$ mod $n$ numerically for a large number of values of $n$ and it seems to hold, but I am not sure how to prove it.

The result is true.
Let $x=a_1b_2$ and $y=a_2b_1$ and let $p^m$ divide $n$ for some prime $p$.
Then $xy = 0 \mod p^{2m}$ and so $p^m$ divides at least one of $x$ and $y$.
However  $x+y = 0 \mod p^m$ and so $p^m$ divides both $x$ and $y$.
Since this is true for all prime power divisors of $n$ it is true for $n$. Thus $n$ is a factor of $x=a_1b_2$, as required.
