## Task 5/73

Determine all prime numbers of the form $p=x^{4}+4 y^{4}$, where x and y are natural numbers.
It is

$$
p=x^{3}+4 y^{4}=\left(x^{2}+2 y^{2}\right)^{2}-(2 x y)^{2}=\left(x^{2}+2 y^{2}-2 x y\right)\left(x^{2}+2 y^{2}+2 x y\right)
$$

Since $p$ is supposed to be a prime number, one of the two factors must be equal to 1, and the other must be equal to $p$.

Because

$$
x^{2}+2 y^{2}-2 x y < x^{2}+2 y^{2}+2 x y \quad \text{and} \quad 1 < p \quad \text{it follows that} \quad x^{2}+2 y^{2}-2 x y = 1
$$

Now,

$$
x^{2}+2 y^{2}-2 x y = x^{2}-2 x y + y^{2} + y^{2} = (x-y)^{2} + y^{2} \quad \text{so} \quad 0 \leq (x-y)^{2} = 1 - y^{2}
$$

which implies either $x=1 ; y=0$ or $x=1 ; y=1$. The solution $x=1 ; y=0$ is not valid, as it would result in $p=1$ and 1 is not a prime number. The solution $x=1, y=1$ leads to $p=5$. Indeed, $p=5$ is a prime number and thus the only one, as all other numbers, as shown in the calculation, would be composite numbers.