<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] $y=\frac{48x}{x-12}$
Now $x-12$ must be multiple of 16, else y will be even.
Let x-12=16k=>$y=\frac{48(16k+12)}{16k}=>\frac{3(16k+12)}{k}=48+\frac{36}{k}$
So, k must divide 36 and must be of the form 4r, where r is an odd integer.
$1≤y≤61=>1≤48+\frac{36}{k}≤61$
$\frac{36}{k}≤13=>k≥3$. 
But $16k=x-12=>x=12+16k>0=>k>-1$
$=>k≥3$
So, the possible values of k are 4,12,36.
$k=4=>y=48+9=57, x=16\cdot 4+12=76.$
$k=12=>y=48+3=51, x=16\cdot 12+12=204$
$k=36=>y=48+1=49, x=16\cdot 36+12=588$
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