I have got about 4 points by trial and error but I guess there are more. Please tell a more systematic approach instead of trial and error and any more points if they exist.
Determine the no. of points $(x,y)$ on the hyperbola $2xy-5x+y=55$ for which $x$ and $y$ are integers
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number-theory
3 Answers
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Write the equation as $$y (2x+1)=5x+55 \Rightarrow y=\frac {5x+55}{2x+1} $$ From this equation it follows that $2x+1$ is a divisor of $5x+55$. Since, $5x+55=2 (2x+1)+(x+53) $, it further follows that $2x+1$ is a divisor of $x+53$ and therefore also of $2 (x+53) $. Since, $2 (x+53)=2x+1+105$, it gives us that $2x+1$ is a divisor of $105$.
Since $105=3\times 5\times 7$, the divisors of $105$ are $\pm 1,\pm 3,\pm 5,\pm 7,\pm 15,\pm 21,\pm 35,\pm 105$. Hope you can take it from here.
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Hint $$x(2y-5)=55-y$$ thus $$x=\frac{55-y}{2y-5}\in \mathbb{Z}$$ we have $2y-5\,\Big |\,55-y$. In other words $2y-5\,\Big |\,105$.
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0Ummm...I don't understand how u can tell "2y-5|55-y. In other words 2y-5|105"? – 2017-01-04
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0$2y-5\,\Big |\,55-y \implies 2y-5\,\Big |\,110-2y$ . On the other hand $2y-5\,\Big |\,2y-5$ thus $2y-5\,\Big |(\,110-2y)+(2y-5)=105$ – 2017-01-04
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Hint: Notice that $$(2x+1)(2y-5) = 4xy - 10x + 2y - 5 = 2(2xy - 5x + y) - 5 = 2(55)-5 = 105. $$ Thus $(2x+1)(2y-5) = 105 = 3\times 5\times 7$. Now, $105$ has $16$ integer factors (why?), so for each such factor $n$, see if the equations $2x+1 = n$ and $2y-5 = \frac{105}{n}$ have integer solutions.
In this case, for each factor the above equations have integer solutions, since $2x+1 = n$ has an integer solution $x$ if and only if $n$ is odd, and similarly for $2y-5$.