Is there an integer $n$ for which the expression $n^{2}+46 n+4441$ is equal to the fourth power of a natural number?
Suppose there is a natural number $N$ for which

$$
n^{2}+46 n+4441=N^{4}
$$

By completing the square for the first two terms on the left side and denoting the resulting square as $P^{2}$,

$$
(n+23)^{2}+3912=P^{2}+3912=N^{4}
$$

From this,

$$
3912=N^{4}-P^{2}=\left(N^{2}-P\right)\left(N^{2}+P\right)
$$

The existence of $P$ and $N$ now depends on whether the number 3912 can be factored into two integers such that the arithmetic mean of the factors - i.e., $N^{2}$ - is a perfect square. Factoring 3912 into prime powers, we get $3912=2^{3} \cdot 3 \cdot 163$. In every two-factor factorization of this, one factor is even. Therefore, it is sufficient to consider factorizations where the other factor is also even, because the arithmetic mean of an even and an odd number is not an integer. Dividing both factors by 2,

$$
\frac{N^{2}-P}{2} \cdot \frac{N^{2}+P}{2}=978=2 \cdot 3 \cdot 163
$$

The two-factor factorizations of 978 and the sums of the factors in these factorizations are:

$$
\begin{array}{rrrr}
978=1 \cdot 978, & 2 \cdot 489, & 3 \cdot 326, & 6 \cdot 163 \\
\text { sum } 979 ; & 491 ; & 329 ; & 169=13^{2}
\end{array}
$$

The last one is suitable. According to this,

$$
\frac{N^{2}-P}{2}=6, \quad \frac{N^{2}+P}{2}=163, \quad N^{2}=169, \quad P=157
$$

Finally, $n+23= \pm P= \pm 157$. According to all of this, there are two integers $n$: $n_{1}=-180$ and $n_{2}=134$, for which the expression $n^{2}+46 n+4441$ is equal to a fourth power of a natural number: $169^{2}=13^{4}$.

István Kerényi (Budapest, Bláthy O. erősáramú ip. t. IV. o. t.)

Remark. We essentially arrive at the same solution if we examine for which $N$ the equation (1) gives an integer root for $n$, since we solved this equation by completing the square instead of using the formula.