$ \frac{2}{x} + \frac{5}{y} = \frac{1}{3}$ the number of solutions of this equation in different cases i.e when x,y are even,odd

Question

$$ \frac{2}{x} + \frac{5}{y} = \frac{1}{3}$$ where $x,y$ belongs to Natural numbers. Prove that it has

• 6 solutions when both $x\ \&\ y$ are even
• 6 solutions when both $x\ \&\ y$ are odd
• 0 solutions when $x$ is even and $y$ is odd
• 0 solutions when $x$ is odd and $y$ is even

I tried doing it case by case for the first case taking $x=2n$ and $y=2m$ and I got $ \frac{5}{2m} = \frac{n-3}{3n}$ from there I tried to find values of $m,n$ such that $x,y$ comes out natural number, I found $(16,24)$. This process is very lengthy and if the question asked to find the number of solutions instead of proving, it would fail miserably.

How can I analyse this to find the solutions?

Answer 1

$ \frac{2}{x} + \frac{5}{y} = \frac{1}{3}$

$ xy-15x-6y=0$

$xy-15x-6y+90=90$

$ (x-6)(y-15)=90$

$ (x-6)(y-15)=1\times 90 = 2\times 45= 3\times 30 = 5\times 18= 6\times 15 = 9\times 10 $

One of integer factors of $90$ is even and the other one is odd. And there are $6$ pairs. Hence there are $6$ solutions when $x$ and $y$ have different parity, no solutions when they have same parity.