Which are the two consecutive odd numbers in the number sequence, the sum of whose squares is of the form $\frac{n(n+1)}{2}$, where $n$ is a natural number.
Let the two adjacent odd numbers be $2k-1$ and $2k+1$, then

$$
(2k-1)^{2}+(2k+1)^{2}=\frac{n(n+1)}{2}
$$

which means

$$
16k^{2}+4=n(n+1)
$$

Multiplying this by 4 and adding 1,

$$
64k^{2}+17=(2n+1)^{2}
$$

Denoting $(2n+1)$ by $m$, we get

$$
m^{2}-64k^{2}=(m+8k)(m-8k)=17
$$

Here, by assumption, $m$ is positive, so the first factor on the left side is also positive, and thus the second factor must also be positive and smaller than the first factor. Since 17 is a prime number, the only possible factorization is

$$
m+8k=17, \quad m-8k=1, \quad \text{and thus} \quad 16k=17-1=16
$$

From this,

$$
k=1, \quad m=2n+1=9, \quad \text{and thus} \quad n=4
$$

Indeed,

$$
1^{2}+3^{2}=\frac{4 \cdot 5}{2}
$$