# $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, 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).

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There is a neat trick for solving such equations in positive integers, which is to note that $x$ must be larger than 12 and $y$ larger than 48, so we can put $x=12+s$ and $y=48+t$ with $s$ and $t$ positive integers and $t$ odd. It looks complicated, but it simplifies through.

More generally, for integer solutions of $$\frac a x + \frac by=\frac 1 z$$ where everything is an integer set $x=az+s$ and $y=bz+t$.

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This is entirely equivalent to Ross Millikan's solution - in the general case we get $st=abz^2$. This applies also when $a=b=1$ which is useful to know if you encounter Egyptian Fractions. –  Mark Bennet Aug 15 '12 at 16:24

You can make it $(x-12)(y-48)=12\cdot 48$ so $y-48$ can be $1,3,9$

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$y=\frac{48x}{x-12}$

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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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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