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I encountered this problem in "Elementary Mathematics" by Dorofeev.

For what natural numbers n is the fraction $\frac{2n+3}{5n+7}$ reducible to lower terms?

So this means that $2n + 3$ and $5n + 7$ must share a common divisor $D$ for it to be reducible. But setting $2n+3=pD$ and $5n+7 = qD$ doesn't really get me anywhere. Where am I going wrong here?

geomquestion
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  • You're not going wrong anywhere, try and show that $\gcd(2n+3, ~5n+7)=1$ – AgentS Mar 13 '15 at 15:42
  • @ganeshie8 Actually, that's a step in the right directon, if one continues by working mod $D$ to eliminate $,n,,$ see the Remark in my answer. – Bill Dubuque Mar 13 '15 at 16:21

3 Answers3

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Hint: you have $$5(2n+3)-2(5n+7)=1$$

Macavity
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The linear map $\rm\,(n,1)\mapsto (5n\!+\!7,2n\!+\!3)\,$ has $\, \det = 5\cdot3-7\cdot 2\!=\color{}{1},\,$ hence, by this very simple theorem, $\, \gcd(5n\!+\!7,2n\!+\!3) =\gcd(n,1)=\color{}{1}$

Remark $\ $ To continue with your approach we can work with linear equations modulo a common divisor $\,d\,$ to eliminate $\,n,\,$ namely, if $\ d\mid 2n\!+\!3,5n\!+\!7\,$ then

${\rm mod}\ d\!:\,\ \begin{align} 2n\equiv -3\\ 5n\equiv -7\end{align}$ $\ \Rightarrow\ \begin{align} 10n\equiv -15\\ 10n\equiv -14\end{align}\ \Rightarrow\ {-}15\equiv -14\ \Rightarrow\ d\mid 15\!-\!14 = 1\,\Rightarrow\,d=1$

The Theorem in the linked answer systematizes this elimination process.

Community
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Bill Dubuque
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$\gcd(5n + 7, 2n + 3) = \gcd(n + 1, 2n + 3) = \gcd(n + 1, 1) = 1$.

filipos
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