Bilateral $mathcal Z$-transform of exponential

We all know that $a^nu(n)$ has unilateral $\mathcal Z$-transform. But what is the $\mathcal Z$-transform of $a^n$? (bilateral) When i tried to solve, i got answer as 'zero'.

But bilateral Laplace transform of $e^t$ doesn't exist. Both are exponentials in discrete and continuous domain respectively. Considering the similarity between Laplace and $\mathcal Z$-transform, how to explain the above problem?

Below, this is how I got 'zero'

$$a^n=a^nu(n) + a^nu(-n-1),$$

Now taking $\mathcal Z$-transform on both sides we get $$\frac{z}{z-a}\quad \text{and}\quad \frac{-z}{z-a}$$ respectively which add to 'zero'

Answer

In complete analogy with the bilateral Laplace transform of $x(t)=e^{-at}$ (which doesn't exist), the bilateral $\mathcal{Z}$-transform of $a^n$ doesn't exist either. The series

$$\sum_{n=-\infty}^{\infty}a^nz^{-n}$$

converges nowhere, simply because $a^n$ grows without bounds for $n\rightarrow -\infty$ if $|a|<1$, or for $n\rightarrow\infty$ if $|a|>1$. Of course, for $|a|=1$ there series doesn't converge either.

EDIT:

As for your computation of the $\mathcal{Z}$-transform of $a^n$, the mistake lies in the fact that in addition to the algebraic expression of the transform you also need to consider the region of convergence. If you split $a^n$ (as you did) as

$$a^n=a^nu[n]+a^nu[-n-1]\tag{1}$$

you can compute the $\mathcal{Z}$-transform of both right-hand side expressions separately:

$$\mathcal{Z}\{a^nu[n]\}=\frac{z}{z-a},\quad |z|>|a|\\ \mathcal{Z}\{a^nu[-n-1]\}=-\frac{z}{z-a},\quad |z|<|a|\tag{2}$$

Note that the region of convergence (ROC) for the first part is outside the circle with radius $|a|$, whereas the ROC of the second part is inside the circle with radius $|a|$. The ROC of the total expression would be the overlap of the two ROCs, which is zero. Consequently, the sum doesn't converge anywhere and the $\mathcal{Z}$-transform of the total expression doesn't exist.