To determine the most general form of three perfect squares that form an arithmetic progression. From this, the three smallest squares that have this property should be derived.

 untranslated portion: (Please note, the last sentence was not translated due to potential ambiguity in the original text. If you could provide more context or clarify, I would be happy to assist further.) 

However, based on the context, a possible translation could be: "From this, derive the three smallest squares that possess this property."
For the later transformations to be simpler, let the three numbers be

$$
n^{2}, \quad(n+2 x)^{2}, \quad(n+2 y)^{2}
$$

According to the condition:

$$
2(n+2 x)^{2}=(n+2 y)^{2}+n^{2}
$$

which means:

$$
(n+2 x)^{2}=(n+y)^{2}+y^{2}
$$

from which:

$$
(n+2 x)=\sqrt{\left[(n+y)^{2}+y^{2}\right]}
$$

$(n+2 x)$ is a rational integer if:

$$
\left[(n+y)^{2}+y^{2}\right]
$$

is a perfect square, that is, if $(n+y)$ and $y$ are so-called Pythagorean numbers. Let:

$$
(n+y)=2 p q
$$

and

$$
y=p^{2}-q^{2}
$$

because then indeed:

$$
(n+2 x)=\left(p^{2}+q^{2}\right)
$$

If the sequence is increasing, which we can assume, then $x$ and $y$ are positive integers, and from (2) it follows that

$$
|p|>|q|
$$

and from (1):

$$
|q|>0
$$

that is: $p$ is at least 2 and $q$ is at least 1.

Based on (1) and (2), the period of the sought numbers is:

$$
\left(q^{2}+2 p q-p^{2}\right)^{2}, \quad\left(p^{2}+q^{2}\right)^{2}, \quad\left(p^{2}+2 p q-q^{2}\right)^{2}
$$

The constant difference of the sequence is $4 p q\left(p^{2}-q^{2}\right)$. This difference is always even, which implies that the terms are simultaneously even or odd. This condition depends on $p$ and $q$. The terms of the sequence are even if $p$ and $q$ are both even or both odd. The terms of the sequence are odd if one of $p$ and $q$ is even and the other is odd. According to the general form, the base number of the second term of the period is always the sum of two perfect squares, and based on this, we get the answer to the second part of the question: which three squares form the smallest arithmetic sequence? In the sequence of numbers, 5 is the smallest number that is the sum of two perfect squares $\left(2^{2}+1^{2}\right)$, so substituting $p=2, q=1$ leads to the desired numbers, that is:

$$
\begin{array}{ccc}
1^{2} & 5^{2} & 7^{2} \\
1 & 25 & 49
\end{array}
$$

Naturally, the periods with the smallest intervals are found when, for odd numbers,

$$
p-q=1 \quad \text { and }
$$

for even numbers, $p-q=2$.

The examination of the general forms shows further that a certain period, determined by substituting $p=4$ and $q=n-1$, if in the next period we substitute $p=n+1$ and $q=n$: the first term of the new period is equal to the last term of the old period. Indeed, for $p=n$ and $q=n-1$, that is, for odd numbers:

$$
\left(2 n^{2}-4 n+1\right)^{2}, \quad\left(2 n^{2}-2 n+1\right)^{2}, \quad\left(2 n^{2}-1\right)^{2}
$$

form the sequence, while for

$$
p=n+1 \quad \text { and } \quad q=n
$$

substitution:

$$
\left(2 n^{2}-1\right)^{2}, \quad\left(2 n^{2}+2 n+1\right)^{2}, \quad\left(2 n^{2}+4 n+1\right)^{2}
$$

will be the terms of the sequence.

For the sequence of even numbers, however:

$$
p=m, \quad q=m-2
$$

for substitution (a):

$$
\left(2 m^{2}-8 m+4\right)^{2}, \quad\left(2 m^{2}-4 m+4\right)^{2}, \quad\left(2 m^{2}-4\right)^{2}
$$

and for

$$
p=m+2, \quad q=m
$$

substitution:

$$
\left(2 m^{2}-4\right)^{2}, \quad\left(2 m^{2}+4 m+4\right)^{2}, \quad\left(2 m^{2}+8 m+4\right)^{2}
$$

will be the terms of the period (b).

In case (b):

$$
\left(2 m^{2}-4 m-2\right)^{2}, \quad\left(2 m^{2}+2\right)^{2}, \quad\left(2 m^{2}+4 m-2\right)^{2}
$$

while for

$$
p=m+3, \quad q=m+1
$$

$$
\left(2 m^{2}+4 m-2\right)^{2}, \quad\left(2 m^{2}+8 m+10\right)^{2}, \quad\left(2 m^{2}+12 m+14\right)^{2}
$$

give the periods.