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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 $ad - bc > 1$.

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For a=2, d=3, b=2, c=1 the statement is not true. Are you sure you wrote it down correctly?

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  • $\begingroup$ I am sorry, I do not understand your question. Is there anyone who could help you state the problem more clearly? $\endgroup$ May 9 '14 at 7:07
  • $\begingroup$ i'm sorry . i want to show that at least one of a,b,c,d is not divisible by ad-bc in your numbers ad-bc = 4 and a is not divisible by ad-bc ! $\endgroup$ May 9 '14 at 7:09

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