<s>[INST] $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y). $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
 [/INST] Let, $y=2n-1$
given, $2n-1\lt61$, $2n \lt 62$
$\frac{1}{x}+\frac{4}{y}=\frac{1}{12}$
$\frac{1}{x}+\frac{4}{2n-1}=\frac{1}{12}$
$\frac{1}{x}=\frac{1}{12}-\frac{4}{2n-1}$
$\frac{1}{x}=\frac{2n-1-48}{(12)(2n-1)}$
$\frac{1}{x}=\frac{2n-49}{(12)(2n-1)}$
$x=\frac{(12)(2n-1)}{2n-49}$
$x=\frac{(1)(3)(4)(2n-1)}{2n-49}$
By careful observation, we see that $2n-49$ needs to be odd as $2n$ is even
i.e., $2n-49=1$, $2n-49=3$, $2n-49=(3)(k)$ where, k is an odd integer
i.e., $n=25$, $n=26$, $2n=49+3k$
i.e., $n=25$, $n=26$, $2n=58$
i.e., $n=25$, $n=26$, $n=29$
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