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.

Bill Dubuque
  • 272,048
Heidar
  • 401

2 Answers2

3

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.

2

Hypotheses $\Rightarrow \dfrac{a_1b_2}n$ & $\dfrac{a_2b_1}n\,$ have sum & product $\in\Bbb Z\,$ thus both are $\in\Bbb Z,\,$ by Rational Root Test

Bill Dubuque
  • 272,048
  • 1
    Thank you. I bookmarked this question since I suspected there's a way to prove it without looking at individual prime factors of $n$. And there is one! – Neat Math Jan 12 '21 at 14:21