Find the number of positive integral solutions of the equation $$\frac{1}{x} + \frac{2}{y} = \frac{1}{4}$$
My attempt:
$$\begin{align} y+2x&=\frac{xy}{4} \\ 4y+8x &= xy \\ 8x &= y(x-4) \\ \end{align}$$
How can i proceed?
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Use SFFT to factor $xy-8x-4y = 0$. – Jul 03 '22 at 17:05
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From $4y+8x=xy$ we have $xy-4y-8x=0\implies(x-4)(y-8)=32$. Note that $-32=1\cdot32, -\cdot16, \dots,-1\cdot-32,-2\cdot-16\dots$ Use this to solve for $x$ and $y$. You will find that there are $12$ solutions in total.
bobeyt6
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You should have $-8x$, which means your equations are slightly off from then on – Milten Jul 03 '22 at 21:30
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 05 '22 at 08:31
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Multiplying by $4xy$ gives $0 = xy - 8x - 4y = (x - 4)(y - 8) - 32$ so that
$$(x - 4)(y - 8) = 32 = 2^{5}.$$
There are $12$ solutions to this equation because there are $12$ solutions to the equation $a + b = 5$ (each of the two factors in the product $2^{5}$ must contribute some number of $2's$, and it must sum to $5$).
Ekesh Kumar
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I’d say there are $6$ (nonnegative integer) solutions to $a+b=5$, and then the number of solutions $x,y$ is twice that. – Milten Jul 03 '22 at 21:28
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 05 '22 at 08:30