133. Find all arithmetic progressions for which the sums of the first $n_{1}$ and $n_{2}\left(n_{1} \neq n_{2}\right)$ terms are equal to $n_{1}^{2}$ and $n_{2}^{2}$, respectively.
S o l u t i o n. By the formula for the sum of the first $n$ terms of an arithmetic progression $S_{n}=\left(a_{1}+\frac{d}{2}(n-1)\right)$, we have:

$$
\begin{aligned}
& \left(a_{1}+\frac{d}{2}\left(n_{1}-1\right)\right) n_{1}=n_{1}^{2} \\
& \left(a_{1}+\frac{d}{2}\left(n_{2}-1\right)\right) n_{2}=n_{2}^{2}
\end{aligned}
$$

from which $a_{1}-\frac{d}{2}=n_{1}\left(1-\frac{d}{2}\right), a_{1}-\frac{d}{2}=n_{2}\left(1-\frac{d}{2}\right)$, i.e.

$$
n_{1}\left(1-\frac{d}{2}\right)=n_{2}\left(1-\frac{d}{2}\right)
$$

thus, $\left(n_{2}-n_{1}\right)\left(1-\frac{d}{2}\right)=0$. Since $n_{1} \neq n_{2}$, then $1-\frac{d}{2}=0, d=2$. From the equality $a_{1}-\frac{d}{2}=0$ we find $a_{1}=1$.

A n s w e r: $1,3,5,7,9, \ldots$.