<s>[INST] show this $\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$ let $a_{1},b_{1},a_{2},b_{2}$ be no-zero integer,and such $\gcd{(a_{i},b_{i})}=1,|a_{i}|\neq 1,|b_{i}|\neq 1,i=1,2$
show that
$$\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$$
My attemp
Assmue that prime $p$ such
$$p|(a_{1}a_{2}-b_{1}b_{2}),p|(a_{1}b_{2}+a_{2}b_{1})$$
It follows that
$$p|[(a_{1}b_{2}+a_{2}b_{1})a_{2}-(a_{1}a_{2}-b_{1}b_{2})b_{2}]$$
$$p|b_{1}(a^2_{2}+b^2_{2})$$
What approaches do you think I could take to solving the next step?
 [/INST] This is not true. For counterexample: $(3+4i)\overbrace{(11+2i)}^{(3-4i)(1+2i)}=25+50i$.
That is, take $\gcd(3,4)=1$ and $\gcd(11,2)=1$. Then
$$
\gcd(3\cdot11-4\cdot2,3\cdot2+4\cdot11)=\gcd(25,50)=25
$$
Another counterexample: $(5+2i)\overbrace{(7+3i)}^{(5-2i)(1+i)}=29+29i$.
That is, take $\gcd(5,2)=1$ and $\gcd(7,3)=1$. Then
$$
\gcd(5\cdot7-2\cdot3,5\cdot3+2\cdot7)=\gcd(29,29)=29
$$
</s>