we have $ad-bc >1$ is it true that at least one of $a,b,c,d$ is not divisible by $ad-bc$ ? Thanks in advance.
Example: $a=2$ , $b = 1$, $c = 2$, $d = 2$, $ad-bc = 2$
so $b$ is not divisible by $ad-bc$
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Yes, that is true.
Hint. Consider a common divisor $q$ of $a, b, c, d$. What does that mean for $q^2$ and $ad - bc$? Now assume that $ad - bc$ is a common divisor of $a, b, c, d$ and apply this to $ad - bc$ itself. This is going to give a contradiction with the assumption that $ad - bc > 1$.