Proof. Divisibility number theory

Question

Prove that no cancellation is possible for $$\frac{a_1 + a_2}{b_1 + b_2}$$ if $a_1 b_2-a_2 b_1=\pm 1$.

I'm new at number theory so if you can be simple it would be great.

Here is what I think:

For there to be no cancellation the gcd of them should be $\pm 1$

$$\gcd((a_1+a_2), (b_1+b_2))=d$$

So we have to prove that $d=\pm 1$. So suppose that $d=\pm 1$.

$$\pm 1|a_1+a_2$$

$$\pm 1|b_1+b_2$$

Then $\pm 1=a_1 b_2-a_2 b_1$ ... Nothing from this

Answer 1

Hint: If $d$ divides $a_1+a_2$ and $b_1+b_2$, then $d$ divides $(a_1+a_2)b_2-(b_1+b_2)a_2$. Expand.