How can I find natural number solutions to $\frac{186-x}{11x+1}=y$?

Question

I'm trying to find solutions to $\frac{186-x}{11x+1}=y$, where $x,y \in \mathbb{N}$. I've been researching Diophantine equations to try and solve this, but everything I've found is in the format $ax + by = c$.

Any help would be much appreciated.

Thank you!

Answer 1

$$11xy+x + y = 186$$ $$121xy +11x + 11y = 2046 $$

Note that $$(11x+1)(11y+1) = 121xy + 11x + 11y + 1$$

Together $$ -1 + (11x+1)(11y+1) = 2046 $$ $$ (11x+1)(11y+1) = 2047= 23 \cdot 89 = 1 \cdot 2047$$

Answer 2

For what positive integer values of $x$ is $y = \frac{186-x}{11x+1}$ a positive integer?

Clearly, $y = \frac{186 - x}{11x + 1} < \frac{186 - x}{11} < \frac{186}{11} = 16 + \frac{10}{11} $. So $y$ must be at most 16. What are the bounds on $x$?

$$1 \le \frac{186 - x}{11x + 1} \le 16$$

$$11x + 1 \le 186 - x \le 176x + 16$$ $$12x + 1 \le 186 \text{ and } 186 \le 177x + 16$$ $$x \le 15 + \frac{5}{12} \text{ and } x \ge \frac{170}{177}$$ $$1 \le x \le 15$$

This is a small enough solution space to brute-force. When we do that, we find two solutions:

• $x = 2, y = 8$
• $x = 8, y = 2$