Hacker News Reader

Another lovely aspect to this problem is that the quintic _is_ solvable if you allow yourself to use certain special functions naturally associated to the icosahedron.

The genesis of the relationship is that the group of rotations of the icosahedron is the same group, A_5.

View on HN

It's solvable in many other ways. There's no rule in algebra that states each individual variable must be a different number. Thus, every variable = 0 then the equation is solved.

No that wouldn't work. Consider the equation x^5 + 1 = 0. Clearly that isn't solvable by x=0.

That's not a "solution" in the algebraic sense that we're talking about.

Reply on news.ycombinator.com