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Find the positive integer solutions for: $\frac{4}{x} + \frac{10}{y} = 1$

I had calculated the solutions manually but it was a very tedious process. Is there any better way to do this?

nonuser
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shayeri bose
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    It's a finite problem, with the possible solutions in a limited range. Simple search sounds like it should be nearly optimal. – lulu Jul 19 '21 at 15:11
  • Perhaps it helps a bit to rearrange to get $y = \frac {10x}{x-4}$ so $(x-4),|,10x$. – lulu Jul 19 '21 at 15:16
  • Extending the comment of @lulu, if $x$ is odd, then $(x - 4)$ is odd, which implies that $(x - 4)$ divides $5x$ which implies that either $(x - 4)$ divides $5$ or $(x-4)$ divides $x$. Similar considerations/conclusions may be present under the hypothesis that $x \equiv 2\pmod{4}$ or $x \equiv 0\pmod{4}$. Anyway, exploring these ideas, while they might not help, would be my first try. – user2661923 Jul 19 '21 at 15:21
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    I'd write it as $(x-4)(y-10)=40$. – Jaap Scherphuis Jul 19 '21 at 15:55
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    Clear denom's then complete the product. – Bill Dubuque Jul 20 '21 at 01:40

1 Answers1

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Let $x\leq y$

$$1=\frac{4}{x}+\frac{10}{y}\leq \frac{4}{x}+\frac{10}{x}=\frac{14}{x}$$

$$x\leq 14$$

$$y=\frac{10x}{x-4}$$

$$5\leq x\leq14$$

$x=5,y=50$

$x=6,y=30$

$x=8,y=20$

$x=9,y=18$

$x=12,y=15$

$x=14,y=14$

Let $y < x$

$$1=\frac{4}{y}+\frac{10}{y}< \frac{4}{y}+\frac{10}{y}=\frac{14}{y}$$

$$y < 14$$

$$x=\frac{4y}{y-10}$$

$$11\leq y\leq13$$

$x=24, y=12$

$x=44, y=11$

Lion Heart
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