Solving for $(x,y): 2+\frac1{x+\frac1{y+\frac15}}=\frac{478}{221}$

Question

Solving for $x,y\in\mathbb{N}$:

$$2+\dfrac1{x+\dfrac1{y+\dfrac15}}=\frac{478}{221}$$

This doesn't make any sense; I made $y+\frac15=\frac{5y+1}5$, and so on, but it turns out to be a very complicated fraction on the left hand side, and I don't even know if what I got is correct.

Is there a more mathematical way to approach this problem?

Answer 1

Yes, there is a very large and important mathematical theory, called the theory of Continued Fractions. These have many uses, both in number theory and in analysis (approximation of functions).

Let's go backwards.

The number $\frac{478}{221}$ is $2+\frac{36}{221}$, which is $2+\frac{1}{\frac{221}{36}}$.

But $\frac{221}{36}=6+\frac{5}{36}$, which is $6+\frac{1}{\frac{36}{5}}$.

Finally, $\frac{36}{5}=7+\frac{1}{5}$.

Now compare the results of these calculations with your expression.

Answer 2

Let's see, $$\frac{478}{221}=2+\frac{36}{221}=2+\dfrac1{\frac{221}{36}}=2+\frac1{6+\frac5{36}}=2+\dfrac1{6+\dfrac1{\frac{36}{5}}}=2+\dfrac1{6+\dfrac1{7+\frac15}}$$

Thus, $x=6,y=7$.

Answer 3

To see uniqueness,

we look at Danny's solution.

Let's see, \begin{align} \frac{478}{221}&=2+\frac{36}{221} &\text{2 is the only integer which will make the leftover fraction less than 1}\\ &=2+\dfrac1{\frac{221}{36}}\\& =2+\frac1{6+\frac5{36}} &\text{Here, 6 is the only number which will make the left over fraction positive and less than 1}\\ &=2+\dfrac1{6+\dfrac1{\frac{36}{5}}}\\ &=2+\dfrac1{6+\dfrac1{7+\frac15}} &\text{and here, 7 is the only number} \end{align}

Thus, $x=6,y=7$.