Find all integer solutions for x and y.

I can solve linear diophantine equations without a problem normally but this has me stumped.

  • 1
    $\begingroup$ In general, you count precisely $$\sum_{x | n}\sum_{y | n/x} 1 = \tau(n^2)$$ Positive integer solutions to $1/x + 1/y = 1/n$. $\endgroup$ – Balarka Sen Mar 28 '14 at 13:29

The equation gives: $14y = xy - 14x$. So solving for $x$ and get: $x = 14 + \frac{196}{y-4}$.

So $x$ is an integer if $y - 4$ divides $196$. We can take it from here....

  • $\begingroup$ Is moving from $14y = xy - 14x$ to $x = 14 + 196/(y-4)$ any easier than the original problem? $\endgroup$ – ColinK Mar 28 '14 at 13:10
  • $\begingroup$ @ColinK: it is much easier because you can see rightaway what y can be. $\endgroup$ – DeepSea Mar 28 '14 at 20:59
  • $\begingroup$ The $y-4$ in the denominator must be $y-14$ $\endgroup$ – ccorn Mar 29 '14 at 14:32

Rewrite as $xy-14x-14y=0$, and then, in an analogue of completing the square, as $(x-14)(y-14)=196$.

So $x-14$ ranges over the divisors of $196$. Since $196=2^2\cdot 7^2$, $196$ has $(2)(3)(3)$ integer divisors, including the negative divisors. That gives $18$ possible values of $x$.


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