# Do I need to check solution while solving an equation?

I am facing a strange problem in solving the following equation for x and y
$$\frac{4x^2+(x^2+y^2-1)^2}{(x^2+(y-1)^2)^2}=1$$ $$4x^2+(x^2+y^2-1)^2=(x^2+(y-1)^2)^2$$ On solving the terms in brackets, we get
$$4y^3-8y^2+4y+4x^2y=0$$ $$4y(y^2-2y+1+x^2)=0$$ $$4y[(y-1)^2+x^2]=0$$ $$y=0 \space or \space (y-1)^2+x^2=0$$ So one solution is $y=0$.
$$\text{if}\space(y-1)^2+x^2=0\,\space \text{then} \space y=1\space \text{and}\space x=0$$ But if I try to put $y=1$, and $x=0$ in original equation, I get $\frac{0}{0}$, which is not $1$.
So I want to ask is $(y=1,$ $x=0)$ solution or not ?

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You could save yourself some manipulation bu using the difference of two squares for your more complicated terms. – Mark Bennet Jun 28 '12 at 7:56

$(x,y)=(0,1)$ is not a solution. \begin{align*} &\frac{4x^2+(x^2+y^2-1)^2}{\left(x^2+(y-1)^2\right)^2}=1\\ \iff&\begin{cases}4x^2+(x^2+y^2-1)^2=\left(x^2+(y-1)^2\right)^2\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\ \iff&\begin{cases}y\left(x^2+(y-1)^2\right)=0\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\ \iff&\begin{cases}y=0\\x\neq0\hbox{ or }y\neq1\end{cases}\iff y=0 \end{align*}

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Thanks very much. I just want to know did you do third step by opening all brackets from second step (squaring required terms), or by some other method ? – Happy Mittal Jun 28 '12 at 8:12
@HappyMittal See Mark Bennet's comment. But the boring expansion is sometimes necessary. You can try to expand this: $4(y+z)^2(z+x)^2(x+y)^2((yz+zx+xy)(1/(y+z)^2+1/(z+x)^2+1/(x+y)^2)-9/4)$. It's from Iran 96 TST. I spent many hours in expanding it! – Frank Science Jun 28 '12 at 8:20