## Task 24/80

It is required to specify a sufficient, $n$-independent condition for the natural number $a$ that guarantees the number $z=n^{8}+a$ is not a prime number for any natural number $n$.
Certainly, here is the translation of the provided text into English, preserving the original formatting:

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$z=n^{8}+a$ is certainly not a prime number if and only if $z$ can be factored into a product of two natural numbers $b ; c>1$:

$$
z=n^{8}+a=b \cdot c
$$

To investigate under what conditions this is possible, we transform $z$:

$$
\begin{aligned}
z=n^{8}+a & =n^{8}+2 n^{4} \sqrt{a}+a-2 n^{4} \sqrt{a}=\left(n^{4}+\sqrt{a}\right)^{2}-\left(n^{2} \sqrt{2 \sqrt{a}}\right)^{2}= \\
& =\left(n^{4}+\sqrt{a}+n^{2} \sqrt{2 \sqrt{a}}\right)\left(n^{4}+\sqrt{a}-n^{2} \sqrt{2 \sqrt{a}}\right)
\end{aligned}
$$

Both factors are certainly natural numbers if and only if both $\sqrt{a}$ and $\sqrt{2 \sqrt{a}}$ are natural numbers, and if the smaller of the two factors is at least 1. We therefore set

$$
\sqrt{2 \sqrt{a}}=k \in N \rightarrow 2 \sqrt{a}=k^{2} \rightarrow a=0.25 k^{4}
$$

For $a$ to be a natural number, $k$ must be even: $k=2 m, m \in N$. Then $a=4 m^{4}$, $\sqrt{2 \sqrt{a}}=2 m$, and consequently

$$
\begin{gathered}
z=n^{8}+a=n^{8}+4 m^{4}=\left(n^{4}+2 m^{2}+2 n^{2} m\right)\left(n^{4}+2 m^{2}-2 n^{2} m\right)= \\
=\left(n^{4}+2 n^{2} m+m^{2}+m^{2}\right)\left(n^{4}-2 n^{2} m+m^{2}+m^{2}\right)=\left[\left(n^{2}+m^{2}\right)^{2}+m^{2}\right]\left[\left(n^{2}-m^{2}\right)^{2}+m^{2}\right]
\end{gathered}
$$

The smaller of the two factors is $\left[\left(n^{2}-m^{2}\right)^{2}+m^{2}\right]$; it is certainly greater than 1 if $m>1$. This leads to the following conclusion:

The number $z=n^{8}+a$ is certainly not a prime number for any natural number $n$ if $a=4 m^{4}$ with $m \in N, m \geq 1$. Since it has not been shown that no other factorization exists, this condition cannot be considered necessary.

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