Solve the equation: $x^2+\frac {1}{x^2}=2^{1-y^2}$. I found the following problem interesting but do not know how to proceed:   

Solve the  equation:   $$x^2+\frac {1}{x^2}=2^{1-y^2}.$$    

Can someone point me in the right direction?

We can proceed with the problem  in the following way:
We know that for any non-negative $a,b ; \quad a+b \geq 2 \sqrt {ab}$. So, $x^2 + \frac {1}{x^2} \geq 2 \sqrt {x^2. \frac {1}{x^2}}=2$. Hence the min. value of the left hand side of the equation is $2$. On the other hand ,the max. value of the 
right hand side is $2.$ So the given equation is equivalent to $x^2 + \frac {1}{x^2}=2$ and $2^{1-y^2}=2$.Hence solving the equations seperately,we get $x=\pm 1$ and $y=0.$
