Five integers form an arithmetic sequence. Whether we take the sum of the cubes of the first four terms, or the sum of the cubes of the last four terms, in both cases we get 16 times the square of the sum of the considered terms. Determine the numbers.
I. Solution. Let's denote the middle number by $c$ and the difference of the arithmetic sequence by $d$. Then the conditions of the problem are:

$$
\begin{gathered}
(c-2 d)^{3}+(c-d)^{3}+c^{3}+(c+d)^{3}= \\
=16(c-2 d+c-d+c+c+d)^{2}=16(2(2 c-d))^{2} \\
(c-d)^{3}+c^{3}+(c+d)^{3}+(c+2 d)^{3}= \\
=16(c-d+c+c+d+c+2 d)^{2}=16(2(2 c+d))^{2}
\end{gathered}
$$

The second equation can be obtained from the first by substituting $-d$ for $d$ (and writing the terms in reverse order), which can be useful for speeding up the transformations. Simplifying equation (1) and dividing by 2:

$$
2 c^{3}-3 c^{2} d+9 c d^{2}-4 d^{3}=128 c^{2}-128 c d+8 d^{2}
$$

Similarly, from (2):

$$
2 c^{3}+3 c^{2} d+9 c d^{2}+4 d^{3}=128 c^{2}+128 c d+8 d^{2}
$$

Subtract the first equation from the second and divide by 2:

$$
3 c^{2} d+4 d^{3}=128 c d
$$

This holds if $d=0$. Then from (1):

$$
4 c^{3}=256 c^{2}, \quad c=0 \quad \text { or } \quad c=64
$$

If $d \neq 0$, then the equation

$$
3 c^{2}-128 c+4 d^{2}=0
$$

must hold. Multiply by 3 and complete the square for the first two terms:

$$
(3 c-64)^{2}+3(2 d)^{2}=2^{12}
$$

Here, if $c$ is an integer, the first term on the left side is not 0, and by assumption, the second term is not 0 either. Let $3 c-64=2^{k} u, 2 d=2^{l} v$, where $u, v$ are odd and $k, l<6$. $k$ and $l$ cannot be different, because then after factoring out an appropriate power of 2, a larger odd number than 1 would remain, and thus the value $2^{12}$ on the right side could not be obtained. If $k=l$, then we get the equation

$$
u^{2}+3 v^{2}=2^{12-2 k}=4^{6-k}
$$

for which we seek a solution with odd $u$ and $v$. It is known that the square of an odd number gives a remainder of 1 when divided by 8. Therefore, the left side of the equation gives a remainder of 4 when divided by 8, so it can only be a power of 2 if it equals 4, i.e., $k=5, u= \pm 1, v= \pm 1$, thus

$$
3 c-64= \pm 32, \quad 2 d= \pm 32
$$

From the first equation, we only get an integer $c$ with the positive sign:

$$
c=32, \quad d= \pm 16
$$

Thus, the sequences that satisfy the conditions of the problem are:
$0, \quad 0, \quad 0, \quad 0, \quad 0$
$64, \quad 64, \quad 64, \quad 64, \quad 64$
$0, \quad 16, \quad 32, \quad 48, \quad 64$
$64, \quad 48, \quad 32, \quad 16, \quad 0$

and by performing the calculations, we find that these indeed satisfy the conditions.

II. Solution. Using the notation from the previous solution, transform the sum of the first and last, and the second and third terms on the left side of equation (1) using the identity $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$. Then the left side becomes:

$$
\begin{gathered}
(2 c-d)\left[(c-2 d)^{2}-(c-2 d)(c+d)+(c+d)^{2}\right]+(2 c-d)\left[(c-d)^{2}-\right. \\
\left.-(c-d) c+c^{2}\right]=(2 c-d)\left(2 c^{2}-2 c d+8 d^{2}\right)
\end{gathered}
$$

Thus, reducing the equation to 0, we can factor out $2(2 c-d)$:

$$
2(2 c-d)\left(c^{2}-c d+4 d^{2}-64 c+32 d\right)=0
$$

Similarly, from equation (2):

$$
2(2 c+d)\left(c^{2}+c d+4 d^{2}-64 c-32 d\right)=0
$$

If $d=2 c$, then the previous equation holds, and the latter becomes:

$$
8 c^{2}(19 c-128)=0
$$

The only integer solution to this is $c=0$, from which $d=0$.

If $d=-2 c$, then the previous equation becomes (4), and we do not get a new solution.

If $d \neq \pm 2 c$, then the following system of equations must hold:

$$
\begin{aligned}
& c^{2}-c d+4 d^{2}-64 c+32 d=0 \\
& c^{2}+c d+4 d^{2}-64 c-32 d=0
\end{aligned}
$$

Taking the half-sum and half-difference of these equations, we get an equivalent system, since the difference and sum of the latter give the original equations:

$$
c^{2}+4 d^{2}-64 c=c(c-64)+4 d^{2}=0, \quad(c-32) d=0
$$

From the latter equation, either $c=32$ or $d=0$ must hold. In the first case, from the previous equation:

$$
4 d^{2}=32^{2} \quad d= \pm 16
$$

In the latter case:

$$
c=0, \quad \text { or } c=64 .
$$

These lead to the arithmetic sequences found in the previous solution.

Remarks. 1. The conditions give two equations for two unknowns, from which a finite number of value pairs for the unknowns result. As the second solution shows, it is not necessary to significantly use the fact that the numbers are integers, except to exclude the sequence

$$
-384 / 19,-128 / 19,128 / 19,384 / 19,640 / 19
$$

and the sequence obtained by reversing it.

2. In the first solution, the equation (3) derived from (1) and (2) was solved by significantly using the fact that the sought numbers are integers. The value pairs $c=32 / 3, d= \pm 16$ that arose during the solution do not only consist of non-integers, but also do not satisfy the other conditions of the problem. This is not surprising, since the solved equation (3) is not equivalent to the system of equations (1) and (2). Therefore, in the first solution, it is necessary to verify that the sequences obtained satisfy all the conditions. The non-integer sequence mentioned in the previous point, however, did not appear in the first solution, which is explained by the fact that there only integer solutions were sought, and to the value pair $c=128 / 19, d= \pm 256 / 19$, the equation (3) does not have an integer solution.