Let's find an increasing sequence consisting of natural numbers, where the sum of any two consecutive terms is equal to the square of their difference.
I. solution. I. We will show that natural number pairs can be constructed that satisfy the required condition for consecutive terms of the desired sequence, and that the larger term of each such pair is equal to the smaller term of another pair. From this, it already follows that starting from a suitable number pair, we can obtain a sequence of any desired number of terms that satisfies the requirement, by taking the larger number as the smaller term of the next pair, forming the larger term, and repeating this procedure until the desired number of terms is reached.

Let the difference between the larger and smaller terms of the pair be $d$, the smaller term be $a$, so the other term is $a+d$; accordingly, $d$ must also be a natural number. According to the requirement,

$$
a+(a+d)=d^{2}
$$

from which

$$
a=\frac{d^{2}-d}{2}=\frac{d(d-1)}{2}
$$

This expression always yields a natural number if we choose $d$ to be a natural number greater than 1, since one of $d$ and its smaller neighbor is even, so their product is divisible by 2; thus, the other term of the pair is also a natural number. (For $d=1$, $a=0$ would result.)

This also shows that infinitely many pairs can be formed that satisfy the requirement for consecutive terms of the sequence. For example, the two terms of the pair formed from $d=5$ are $a=5 \cdot 4 / 2=10$ and $a+d=15$.

Furthermore, the larger term of the pair formed from an arbitrary $d$ (1) is

$$
a+d=\frac{d^{2}+d}{2}=\frac{(d+1) d}{2}
$$

and this indeed follows from (1) if we substitute $d+1$ for $d$. Accordingly: the larger term of the pair with difference $d$ is equal to the smaller term of the pair with difference $d+1$. The 15 obtained in the example also results from the form $6 \cdot 5 / 2$.

Based on all this, the second term $c_{2}$ of the sequence and the first term $c_{1}$, with the difference $d$ between them (where $d \geq 2$, a natural number), $c_{1}$ is given by (1), $c_{2}=c_{1}+d$, and furthermore, from term to term, we take the number one greater than the difference between the neighboring terms. For example, from $d=5$,

$$
c_{1}=10, \quad c_{2}=10+5=15, \quad c_{3}=15+6=21, \quad c_{4}=21+7=28, \quad \ldots
$$

II. A formula can be given from which the $n$-th term of the sequence can be calculated without generating the preceding terms. The simplest is to choose the smallest allowed value of $d=2$ for $c_{2}-c_{1}$: by swapping the factors in (1),

$$
c_{1}=\frac{1 \cdot 2}{2}=1, \quad c_{2}=\frac{2 \cdot 3}{2}=3, \quad \text { generally } \quad c_{n}=\frac{n(n+1)}{2}
$$

If, however, $c_{2}-c_{1}=d_{1}$, then

$$
c_{1}=\frac{\left(d_{1}-1\right) d_{1}}{2}, \quad c_{2}=\frac{d_{1}\left(d_{1}+1\right)}{2}, \quad \text { generally } \quad c_{n}=\frac{\left(d_{1}+n-2\right)\left(d_{1}+n-1\right)}{2}
$$

Vargha Ágnes (Kisvárda, Császy L. Gymn., II. o. t.)

II. solution. Let the two consecutive terms of the sequence be $u$ and $v$, where $v>u>0$, integers. Then

$$
u+v=(v-u)^{2}
$$

Rearranging in terms of $v$,

$$
v^{2}-(2 u+1) v+\left(u^{2}-u\right)=0
$$

and using the quadratic formula,

$$
v=u+\frac{1}{2}(1+\sqrt{8 u+1})
$$

(The minus sign cannot precede the square root due to the agreement $v>u$.)

This will only be a positive integer for such $u$ for which $8 u+1$ is a perfect square, and indeed a perfect square of an odd number:

$$
8 u+1=(2 k-1)^{2}
$$

where $k$ is an integer. Without loss of generality, we can assume that $k$ is positive, and since $u \geq 1$, $k \geq 2$. Thus, from (4), and then from (3),

$$
\begin{gathered}
u=\frac{(2 k-1)^{2}-1}{8}=\frac{k(k-1)}{2} \\
v=u+k=\frac{k(k+1)}{2}
\end{gathered}
$$

and these are identical to the results (1) and (2) of the first solution.

Sailer Kornél (Ózd, József A. Gymn., II. o. t.)

III. solution. Let the terms of the sequence be $a_{0}, a_{1}, a_{2}, \ldots$ and let the difference between two consecutive terms be $a_{i+1}-a_{i}=$ $b_{i}(i=0,1,2, \ldots)$, which is also a natural number by assumption. The requirement of the problem for two consecutive pairs is:

$$
\begin{aligned}
a_{i+1}+a_{i} & =b_{i}^{2} \\
a_{i+2}+a_{i+1} & =b_{i+1}^{2}
\end{aligned}
$$

The difference of the left sides is

$$
a_{i+2}-a_{i}=\left(a_{i+2}-a_{i+1}\right)+\left(a_{i+1}-a_{i}\right)=b_{i+1}+b_{i}
$$

so the difference of (6) and (5) is

$$
b_{i+1}+b_{i}=b_{i+1}^{2}-b_{i}^{2}
$$

and since the left side is positive, we can simplify with it:

$$
1=b_{i+1}-b_{i}
$$

that is, the increase from term to term increases by 1 from pair to pair, forming an arithmetic sequence:

$$
b_{i}=b_{0}+i
$$

Now, from (5),

$$
\begin{gathered}
\left(a_{i}+b_{i}\right)+a_{i}=b_{i}^{2} \\
a_{i}=\frac{b_{i}^{2}-b_{i}}{2}=\frac{b_{i}\left(b_{i}-1\right)}{2}=\frac{\left(b_{0}+i\right)\left(b_{0}+i-1\right)}{2}
\end{gathered}
$$

and since $a_{0}>0$, $b_{0}>1$.

Conversely, (5) follows from (8) and (7), and also that as the half of the product of two consecutive natural numbers, it is a natural number. Thus, the sequences that satisfy the problem are uniquely determined by the difference of their first two terms $b_{0}$.