Discover which positive integers $x$ and $y$ satisfy the equation $x^{4}=y^{2}+71$.

## List 3
The equation can be written in the form $x^{4}-y^{2}=71$. Now, factoring $x^{4}-y^{2}$ we have:

$$
\left(x^{2}-y\right)\left(x^{2}+y\right)=71(*)
$$

Since $x$ and $y$ are integers, each of the factors $\left(x^{2}-y\right)$ and $\left(x^{2}+y\right)$ is also an integer. Therefore, in $\left(^{*}\right)$, we write 71 as the product of two integers. Since 71 is a prime number, it can only be written as the product of integers in the form: $71=1 \times 71$. Thus, we have two cases to consider: $\left(x^{2}-y\right)=1$ and $\left(x^{2}+y\right)=71$, or $\left(x^{2}-y\right)=71$ and $\left(x^{2}+y\right)=1$.

Let's study each case.
1st case: $\left\{\begin{array}{l}x^{2}-y=1 \\ x^{2}+y=71\end{array}\right.$.

Adding the two equations, we get: $2 x^{2}=72$, which implies $x= \pm 6$. Therefore, $y=( \pm 6)^{2}-1=35$. Since $x, y$ are positive integers, we conclude that the solution in this first case is: $x=6$ and $y=35$.

2nd case: $\left\{\begin{array}{l}x^{2}-y=71 \\ x^{2}+y=1\end{array}\right.$.

If $x^{2}+y=1$, then $x=0$ and $y=1$ or $x=1$ and $y=0$ since $x, y$ are positive integers. On the other hand, it is easy to verify that these values do not satisfy the equation $x^{4}=y^{2}+71$.

Therefore, the solution to the problem is: $x=6$ and $y=35$.