Better understanding of the meaning of implication in equations While solving this equation
$$\sqrt{x^2+1}+x=1$$
I started to have some doubts about my real logical understanding of what I'm doing: I will show the steps to try to communicate better what I mean. We know that adding/subtracting functions that belong to the domain of the original equation gives an equivalent equation (meaning it has the same solutions of the first one), so if I want to add $-x$ both sides the correct logical operator is
$$\sqrt{x^2+1}+x=1\iff\sqrt{x^2+1}=1-x $$
Going on, I know that is true that $a=b\implies a^2=b^2$, so the next step is not an equivalence but an implication; meaning that the correct logic operator is
$$\sqrt{x^2+1}=1-x \implies x^2+1=(1-x)^2$$
Now since $(1-x)^2=x^2-2x+1$, I suppose that an equality is an equivalence (correct me if I'm wrong) so I get
$$x^2+1=(1-x)^2 \iff x^2+1=x^2-2x+1 \iff -2x=0 \iff x=0$$
So, since they are all equivalence and there is an implication, connecting the first and the last I get
$$\sqrt{x^2+1}=1-x \implies x=0$$
So the questions are:
a) since an equation is the problem to determine which unknowns, substituted in the equality, makes the equality true, it seems to me, of course wrongly, that we've obtained the converse, that is "if equation is zero then the unknowns are $*$" while, if I've not understood wrong, we are searching for what unknowns make the equation verified, so we're searching for the converse implication "if unknowns are $*$ these then the equation is verified". What am I missing here?
b) In the equation $\sqrt{x^2+1}=1-x$ I have on the left hand side a non negative quantity while on the right hand side a quantity that can be negative for some values of $x$; so, when arriving to $(2)$, I should add a condition $1-x \geq 0$ to have a real equivalence?
Thanks.
 A: There are two ways of dealing with this. One of them is to do what you did: to be aware of the fact that, at a certain step, what we have is an implication which is not an equivalence. So, at the end, when we get $x=0$, we have to check whether or not $0$ is really a solution of the original equation (it is).
The other way has to do with what you wrote in the final paragraph: when we have $\sqrt{x^2+1}=1-x$ we decide that, from now when, we will work only on $(-\infty,1]$; if $x>1$, then the equation has no solutions (since $\sqrt{x^2+1}\geqslant0$ and $1-x<0$. And, on that interval, we do have an equivalence:$$\sqrt{x^2+1}=1-x\iff x^2+1=(1-x)^2.$$At the end of the computations, we must eliminat all solutions that we got that do not belong to $(-\infty,1]$ (there will be none in this case).
Finally, I dont' see why is that that you assert that what we were supposed to be doing is “if unknowns are ∗ these then the equation is verified”; it's the other way around.
A: 
Since an equation is the problem to determine which unknowns,
substituted in the equality, makes the equality true, it seems to me,
of course wrongly, that we've obtained the converse.

Nope. Nothing wrong with that. It just has to be presented properly.
Theorem: $\sqrt{x^2+1}=1-x \iff x=0$
Proof: You have already proven that $\sqrt{x^2+1}=1-x \implies x=0$. Now, it is trivial to prove the converse: $$x=0 \implies (\sqrt{x^2+1}= 1 ~~\land~~ 1-x=1)\implies \sqrt{x^2+1}=1-x ~~\square$$
