Macedonian Mathematical Competitions

If $p, q, r$ are consecutive terms in the arithmetic progression, then $p + r = 2q$. Because of this, we have
$$\sqrt{y^2 + 2y + 1} + y - 1 = 2 \frac{y^2 + 3y - 1}{3}$$
$$3(\sqrt{(y+1)^2} + y - 1) = 2(y^2 + 3y - 1)$$

Since $\sqrt{(y+1)^2} = |y+1|$, we have
$$\begin{aligned} 3(|y+1| + y - 1) &= 2(y^2 + 3y - 1) \\ 2y^2 + 3y - 3 &= |y+1| + 1 = 0 \end{aligned}$$
The last equation is equivalent with
$$\begin{cases} y + 1 \ge 0 \\ y^2 + 3y - 3(y+1) + 1 = 0 \end{cases}$$
$$\begin{cases} y + 1 < 0 \\ y^2 + 3y + 3(y+1) + 1 = 0 \end{cases}$$
$$\begin{cases} y \ge -1 \\ y^2 - 1 = 0 \end{cases} \text{ or } \begin{cases} y < -1 \\ y^2 + 3y + 2 = 0 \end{cases}$$
The solution of the first system is $y = 1$ or $y = -1$, and for the second one is $y = -2$. Finally $y \in \{-2, -1, 1\}$.