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I am struggling to solve this problem and would appreciate any help:

When is $\frac{12x+5}{12y+2}$ NOT in lowest terms? ($x$,$y$ are nonnegative integers)

I have found that it is not in lowest terms for $x=6$ and $y=9$ because numerator and denominator are divisible by $11$, but I'm stuck here.


EDIT: Apparently "lowest terms" isn't in common usage in maths, so I will have to explain what it means. A fraction $p/q$ with $p,q\in \mathbb{Z}$ and $q\ne 0$ is in lowest terms when $\gcd(p,q)=1$. Otherwise, it is not in lowest terms.

For example, $\frac{3}{5}$ and $\frac{9}{2}$ are in lowest terms, but $\frac{15}{3}$ and $\frac{17}{34}$ are not.

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  • $\begingroup$ Yeah, $\frac{17}{34}$ is a good example. $\endgroup$
    – BaronVT
    Commented Feb 13, 2015 at 19:54
  • $\begingroup$ For another example where it's not in its lowest terms have $x=5$, $y=2$ then numerator and denominator are divisible by 13 $\endgroup$ Commented Feb 13, 2015 at 21:43
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    $\begingroup$ Do you know if the question has a simple answer? There are a lot of solutions, but no pattern is evident. $\endgroup$
    – jjagmath
    Commented Mar 29, 2021 at 3:09
  • $\begingroup$ Judging from the several deleted answers below, there is probably no simple answer. $\endgroup$
    – Toby Mak
    Commented Mar 29, 2021 at 3:42

2 Answers 2

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When is $\dfrac{12x+5}{12y+2} $ irreducible?

This is really just a partial answer.

So, when does $12x + 5 = A$ and $12y + 2 = B$ where $\gcd(A, B)=1$

Note that

\begin{align} 12 &\mid 2A - 5B \\ 5B &\equiv 2A \pmod{12} \\ B &\equiv 10A \pmod{12} \\ B &= 12n + 10A \end{align}

So $\gcd(A,B)=1$ becomes $\gcd(12n, A) = 1$.

\begin{align} 12y + 2 &= B \\ 12y + 2 &= 12n + 10A \\ 12y &= 12n + 10A - 2 \\ \hline 10A &\equiv 2 \pmod{12} \\ 5A &\equiv 1 \pmod 6 \\ A &\equiv - 1 \pmod 6 \\ A &= 6\alpha - 1 \\ \hline 12y &= 12n + 10(6\alpha - 1) - 2 \\ 12y &= 12n + 60\alpha - 12 \\ y &= n + 5\alpha - 1 \\ \hline B &= 12n + 60\alpha - 10 \end{align}

Finally, we solve for $\alpha$.

\begin{align} 12x + 5 &= A \\ 12x &= 6\alpha - 6 \\ 2x &= \alpha - 1 \\ \alpha &= 2x + 1 \end{align}


Pick any integer value for $x$.

$ \alpha = 2x + 1 $

$ A = 12x + 5 $

Pick any $n$ such that $\gcd(12n, A) = 1$.

$B = 12n + 10A$

$y = n + 10x + 4$ implies

$$ \dfrac{12x+5}{12y+2} = \dfrac{12x+5}{12n + 120x + 50}$$

which is irreducible as long as $\gcd(12n, 12x+5) = 1$.

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Fractions in general are in lowest terms $6/\pi^2$ of the time.
One quarter of fractions can cancel 2, one ninth can cancel 3, and so on, so the proportion with no prime factor to cancel is $$\frac34\cdot\frac89\cdot\frac{24}{25}\cdot\frac{48}{49}\cdots=\frac6{\pi^2}$$
For these fractions, $2$ and $3$ will never cancel, so we lose the factors of $3/4$ and $8/9$ from the left-hand side.
These fractions are in lowest terms this proportion of the time: $$\frac43\frac98\frac6{\pi^2}=\frac9{\pi^2}\approx 0.91189$$

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