Prove/Disprove a condition about gcd of numbers

Question

Statement: Let $k,l\in\mathbb{Z}$ be such that $\gcd(k,l)=1$. Then, for all $a,b,c,d\in\mathbb{Z}$ such that $ad-bc=\pm1$, we have $\gcd(ka+lc,kb+ld)=1$.
Prove or disprove this statement.
I have been trying to prove the statement using contradiction. I assumed that $\gcd(ka+lc,kb+ld)=g>1$. I got that none of $a,b,c,d,k,l$ is divisible by $g$. I could not get anything more.

Answer 1

Hint: Set $g = \gcd(ka + lc, kb + ld)$. Then $g$ divides any linear combination of $ka + lc$ and $kb + ld$, so for example $$g\mid d(ka + lc)-c(kb + ld)$$

Can you finish from here?