Powers, powers, ugh! (Part 2)

$y^2+y^3 + \ldots+y^{60} + \ldots+y^{99} = 0\\ y^2(1+y+y^2+y^3+\ldots+y^{94}+y^{95}+y^{96}+y^{97}) = 0\\ y^2\left(1(y+1) + y^2(y+1) + \ldots +y^{94}(y+1) + y^{96}(y+1)\right) = 0\\ y^2(y+1)(1+y^2+\ldots+y^{94} + y^{96}) = 0$

The first factor:

$y^2 = 0 \implies y = 0$

This is not non-zero, so we ignore it.

The second factor:

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

The third factor

$1+y^2+\ldots+y^{94}+y^{96} = 0$

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

This implies that $y^{2} \geq 0, \ldots,\;y^{94} \geq 0,\;y^{96} \geq 0$

Therefore,

$y^2+ \ldots +y^{94} + y^{96} \geq 0\\ 1+y^2+ \ldots+ y^{94} + y^{96} \geq 1$

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

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