A tricky but silly doubt regarding the solutions of $x^2/(y-1)^2=1$ Motivation : 
I have been confused with some degree 2 equation. I suddenly came across a simple equation and couldn't get the quintessence behind that. 
I have an equation $$\dfrac{x^2}{(y-1)^2}=1 \tag{1}$$ and I was looking for its solutions. It was asked by some kid ( of $9$nth standard ) to me. 
I did some manipulation and got $$x^2=(y-1)^2 \tag{2} $$ finally. And one can see that $(0,1)$ satisfies the Equation $[2]$ well. But I was happy, and within small time, 
I realized that the same solution set can't satisfy the equation $[1]$ . If you substitute $(0,1)$ in $[1]$ you get $\dfrac{0}{0}=1$ which is wrong.
The  answer that convinced me finally : 
We can see the same equation as this $x^2. \dfrac{1}{(y-1)^2}=1$ . We know that the set  of integers form a ring. So the product of two numbers is one if one number is the inverse of other number. '$1$' present on the R.H.S is the identity element. So the product of the entity with its inverse always gives us the identity. 
So when $x$ is $0$, the $0$ doesn't have an inverse in the integers. So the case is to be emitted. 
Still persisting questions :
But the thing that makes me surprise is that the Wolfram Alpha gives me this solution . 
                   
In the picture you can clearly see that they both intersect at $(0,1)$ . But what is that confusion ? 
We omitted that solution, but in fact $(0,1)$ is the intersection of the two lines. 
Questions that are to be answered by learned people : 


*

*What is the value of term $\dfrac{0}{0}$ ? Isn't it $1$ ? 

*Why  the solution pair $(0,1)$  satisfies $x^2=(y-1)^2$ but not $\dfrac{x^2}{(y-1)^2}=1$ ? We know that both of them are manifestations of each other in a simple manner. 

*If we need to omit that solution, why do the lines intersect at $(0,1)$ ? 
Thank you everyone for giving your time. 
 A: The equations $$x^2=(y-1)^2\tag{1}$$ and $$\frac{x^2}{(y-1)^2}=1\tag{2}$$ do not have the same solution set. Every solution of $(2)$ is a solution of $(1)$, but $\langle 0,1\rangle$ is a solution of $(1)$ that is not a solution of $(2)$, because $\frac00$ is undefined.
The reason is that $(1)$ does not imply $(2)$. Note first that $(2)$ does imply $(1)$, because you can multiply both sides of $(2)$ by $(y-1)^2$ to get $(1)$. In order to derive $(2)$ from $(1)$, however, you must divide both sides of $(1)$ by $(y-1)^2$, and this is permissible if and only if $(y-1)^2\ne 0$. Thus, $(1)$ and $(2)$ are equivalent if and only if $(y-1)^2\ne 0$. As long as $(y-1)^2\ne 0$, $(1)$ and $(2)$ have exactly the same solutions, but a solution of $(1)$ with $(y-1)^2=0$ need not be (and in fact isn’t) a solution of $(2)$.
As far as the graphs go, the solution of $(1)$ is the union of the straight lines $y=x+1$ and $y=-x+1$. The solution of $(2)$ consists of every point on these two straight lines except their point of intersection.
A: Wolfram Alpha sometimes does that. It's a little bit finicky. You can see that if we ask it for the domain of $z=\frac{x^2}{(y-1)^2}$, it does indeed have us omit the case where $y=0$. I've observed this behaviour lots—for instance if you ask it to plot $y=\frac{(x-1)^2}{x-1}$, you get a similar result: it fills in the hole which should exist at $x=1$. However when you ask it for the domain of that function, it gives the correct domain. I think that this is evidence of a tradeoff between having something which interprets natural language, and having something that's precise.
A: You cannot multiply both side of an equation with $0$. When you multiply your equation by $(y-1)^2$, you are assuring that you are not taking $y=1$ as a valid solution.
A: Two points which might help. 


*

*$0/0$ is undefined

*you are not allowed to multiply an equation by 0 (if you do so then you increase the set of solutions)


So [1] is equivalent to [2] iff $y\neq 1$. 
The solutions to your equation [1] thus read
$$ x =\pm (y-1), \qquad y\neq1.$$
