TITLE: Fake proof that $1$ is the solution of $x^2+x+1=0$
QUESTION [12 upvotes]: So I have this false proof and I am honestly confused why this is happening.
Consider $x^2+x+1=0$, then $x+1=-x^2$.
Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?
I think this might be a really silly question but it is stomping me :P

REPLY [2 votes]: LOGIC. Or maybe just GRAMMAR. $ \;x^2+x+1=0\iff$ $ (x\ne 1\land x^2+x+1=0)\iff$ $ (x\ne 1\land -x^2+1/x=0)\iff$ $ (x\ne 1 \land x^3=1).$
What you have shown is that $x^2+x+1=0\implies x^3=1$ but you have NOT shown 
 that $ x^2+x+1=0\iff x^3=1.$
SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..  
If you omit the "$\implies$" or "$\iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $A\implies B$ or $B\implies A$  or $A\iff B.$