The general solution of $y(n+1) = ay(n)^2$ I would like to find the general solution of the difference equation $y(n+1) = \alpha y(n)^2 $. I know that the general solution to $y(n+1) = y(n)^2$ is $y(n) = \exp({c\cdot 2^{n}})$. However, I've not yet been able to rewrite it to a general solution of $y(n+1) = \alpha y(n)^2$. Is this a efficient approach to finding the general solution. If so, what steps do I need to take to get to the general solution of $y(n+1) = \alpha y(n)^2 $. If not, what would be a better approach to solve this difference equation?
 A: $$
y(n)\rightarrow \beta \bar{y}(n)
$$
thus
$$
\beta \bar{y}(n+1) = \beta^2\alpha\bar{y}(n)^2\implies \bar{y}(n+1) = \alpha\beta\bar{y}(n)^2
$$
let $\alpha\beta = 1\implies \beta = \frac{1}{\alpha}$ thus
$$
\bar{y}(n+1) = \bar{y}(n)^2\implies \bar{y}(n) = \mathrm{e}^{c2^n}
$$
so
$$
y(n) = \frac{1}{\beta}\mathrm{e}^{c2^n}  = \alpha \mathrm{e}^{c2^n}
$$
$\textbf{edit}$
As pointed out by the OP..i made a mistake in the last line.
$$
y(n) = \beta \bar{y}(n) = \beta \mathrm{e}^{c2^n} = \frac{1}{\alpha}\mathrm{e}^{c2^n} 
$$
A: Let $z(n)=\alpha\,y(n)$. Then
$$
z(n+1)=\alpha\,y(n+1)=\alpha^2\,y(n)^2=z(n)^2.
$$
A: I like this methods:
since
$$y_{n+1}=a(y_{n})^2\Longrightarrow \ln{(y_{n+1})}=\ln{a}+2\ln{(y_{n})}$$
let
$$\ln{(y_{n})}=x_{n}$$
so
$$x_{n+1}=\ln{a}+2x_{n}\Longrightarrow  x_{n+1}+\ln{a}=2\left(x_{n}+\ln{a}\right)$$
so
$$\Longrightarrow x_{n}+\ln{a}=(\ln{x_{1}}+\ln{a})2^{n-1}$$
so
$$y_{n}=e^{x_{n}}=e^{(\ln{x_{1}}+\ln{a})2^{n-1}-\ln{a}}=e^{\ln{\left(\dfrac{(ax_{1})^{2^{n-1}}}{a}\right)}}=\dfrac{(ax_{1})^{2^{n-1}}}{a}$$
where $x_{1}=\ln{y_{1}}$
