Powers, powers, ugh!

$y + y^2 + y^3 + \ldots + y^{60} + y^{61} = -1\\ y^{61} + y^{60} + y^{59} + y^{58} + \ldots + y^3 + y^2 + y + 1 = 0\\ y^{60} (y+1) +y^{58}(y+1)+ \ldots + y^2(y+1) + 1(y+1) = 0\\ (y+1)(y^{60} + y^{58} + \ldots + y^2 + 1) = 0$

The first factor:

$y+1 = 0 \implies y=-1$

The second factor:

$y^{60} + y^{58} + \ldots + y^2 + 1 = 0$

Notice that for even positive integers of $n$ , $y^n \geq 0$ for all real values of $y$

This implies that $y^{60} \geq 0,\;y^{58} \geq 0, \ldots,\;y^2 \geq 0$

Therefore,

$y^{60} + y^{58} + \ldots + y^2 \geq 0\\ y^{60} + y^{58} + \ldots + y^2 + 1 \geq 1$

From this, we can conclude that the second factor has no real roots.

Therefore, the only real solution is $y=\boxed{-1}$

A number which doesn't change with its power, it is possible only when the number is 1. But according to the question the power on the number are even & odd number and the RHS is -1.(-1)² = 1 & (-1)³ = -1 . Like this all the evens & odds will cancel each other and the number which is left is (-1)/\61 = -1.