## Task 32/74

For which natural numbers $n$ is $z=4^{n}+4^{4}+4^{8}$ a perfect square?
The number $z=4^{n}+4^{4}+4^{8}$ is a perfect square if and only if the number $z^{\prime}=4^{k}+4+1=\left(2^{k}\right)^{2}+2^{8}+1$ with $k=n-4$ is a perfect square. This is because $z=4^{4} \cdot z^{\prime}=16^{2} \cdot z^{\prime}$. The approach

$$
z^{\prime}=\left(2^{k}\right)^{2}+2^{8}+1=\left(2^{k}\right)^{2}+2 \cdot 2^{7}+1=\left(2^{k}+1\right)^{2}
$$

yields $k=7, n=11$ as a solution. Now, it needs to be checked whether another approach can yield further solutions.

Assume there is an $m$ such that

$$
\left(2^{k}\right)^{2}+2^{8}+1=m ; \quad k \neq 7
$$

then $m$ must be odd: $m=2 r+1$, where $r$ is a natural number. Therefore,

$$
\left(2^{k}\right)^{2}+2^{8}+1=(2 r+1)^{2}=4 r+4 r+1 ; \quad\left(2^{k-1}\right)^{2}+2^{6}=r(r+1)
$$

The right side of the equation is a product of two consecutive numbers, so the left side must also be expressible as such a product. It is easily verified by calculation that this is not possible for $k \leq 4$. For $k>4$ it follows that

$$
\left(2^{k-1}\right)^{2}+2^{6}=2^{6}\left(2^{2 k-8}+1\right) ; \quad \text { thus } \quad r=2^{6}=2^{2 k-8} ; k=7
$$

contradicting the assumption. Therefore, $n=11$ is the only number for which $z$ is a perfect square.