How to prove that if $a$, $b$, $c$, and $d$ are integers divisible by $ab-cd$, then $|ab - cd| = 1$

Question

Have a very good day!

This problem was mentioned in Mathematical Circles: (Russian Experience) and I was trying to solve it but eventually was unavailable.

If $a,b,c$ & $d$ are integers, &

$$ab - cd \mid a, b, c, d$$

Then prove $|ab - cd|=1$

Answer 1

Assuming that $ab-cd\ne0$, we have the following.

From $ab-cd\mid a$, $b$, we get $ab-cd\mid ab$. Let $k(ab-cd)=ab$, where $k\in\Bbb Z$, then $(k-1)ab=kcd$. Since $(k-1,k)=1$, $k-1\mid cd$ and $k\mid ab$. Let $ab=tk$ where $t\in\Bbb Z$, so $cd=t(k-1)$, and $ab-cd=t$. Let $a=At$ and $b=Bt$ where $A$, $B\in\Bbb Z$.

By the same reason, we can let $c=Ct$ and $d=Dt$ where $C$, $D\in\Bbb Z$. Hence, \[ABt^2-CDt^2=t\implies(AB-CD)t=1.\] We may only have $t=\pm1$, meaning that $t=\pm1$ so $|ab-cd|=|t|=1$, as desired.

Answer 2

Let $p$ a prime such that $p \mid ab-cd \mid a,b,c,d$. Let $n\in \mathbb{N}$ be the largest exponent of $p$ that divides $ab-cd$. So $p^n\mid ab-cd$. Then $p^n \mid a,b,c,d \implies p^{2n}\mid ab$.

Can you finish?