Discover which positive integers $x$ and $y$ satisfy the equation $x^{4}=y^{2}+71$.
The equation can be written in the form $x^{4}-y^{2}=71e$, factoring $x^{4}-y^{2}=\left(x^{2}-y\right)\left(x^{2}+y\right)$, in the form

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

Since $x$ and $y$ are integers, each of the factors $x^{2}-y$ and $x^{2}+y$ is also an integer, so we can 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 $1 \times 71$ or $71 \times 1$. Thus, we only have two cases to consider, namely $x^{2}-y=71$ and $x^{2}+y=1$, or $x^{2}-y=1$ and $x^{2}+y=71$. Since $x, y$ are positive integers, we have $x^{4}=y^{2}+71 \geq 72>16=2^{4}$, so $x>2$. In particular, $x^{2}+y=1$ is impossible, as it would imply $1=x^{2}+y \geq 9+1=10$.

Thus, we are left to consider the case $x^{2}-y=1$ and $x^{2}+y=71$. Adding these two equations, we get $2 x^{2}=72$, which gives $x= \pm 6$ and, therefore, $y=( \pm 6)^{2}-1=35$. Since $x, y$ are positive integers, we conclude that the only solution is $x=6$ and $y=35$.