I came across the following question:
The causal discrete-time signal $x[n]$ satisfies: $$x[n]\ast x[n]=u[n]$$ Where $\ast$ is the convolution operator and $u[n]$ is the discrete unit step function. Knowing that $x[0]>0$, compute $x[3]$.
My first attempt, from intuition was to apply the Z-transform to the equation, but I get the following strange thing:
\begin{align} \mathcal{Z}\left\{x[n]\ast x[n]\right\}&=\mathcal{Z}\left\{u[n]\right\}\\ X[z]^2&=\frac{1}{1-z^{-1}}\\ X[z]&=\frac{1}{\sqrt{1-z^{-1}}} \end{align}
Which is not something I have seen before on Z-transform tables. I could probably try to apply the inverse Z-transform from the definition, but I am not sure I would get somewhere.
Also, this question comes from a test where the idea is to be able to give straightforward answers, avoiding computations as much as possible, given the time constraints.
Is there another way around this?
Let $y[n] \triangleq x[n]*x[n],n\geq 0$, with $x[n]= 0,n< 0$ (causal sequence). By definition of the convolution operation, it holds
$$ \begin{align} y[0] &= (x[0])^2 \\ y[1] &= 2x[0]x[1] \\ y[2] &= 2x[0]x[2]+(x[1])^2\\ y[3] &= 2(x[0]x[3]+x[1]x[2]) \end{align} $$
By assumption, $y[n]=1, n\geq 0,$ and $x[0]\neq 0$, and you can solve these equations to find $x[3]$.