= Solution
The independent <Laplace distribution> responses have log likelihood
$$
\ell(\beta,\sigma)=-n\log(2\sigma)-\frac{S_1(\beta)}\sigma.
$$
For every fixed positive $\sigma$, maximizing this expression means minimizing the sum of absolute residuals. Thus \b[$\widehat\beta$ is a <least absolute deviations> estimate], not a least-squares estimate. Such a minimizer exists: minimize the distance in the finite-dimensional closed column space of the design matrix. It need not be unique if the data or design allow multiple minimizers.
Write $s=S_1(\widehat\beta)$. For $s>0$, differentiating the profiled log likelihood gives $-n/\sigma+s/\sigma^2=0$. Its unique maximum is
$$
\boxed{\widehat\sigma=\frac{S_1(\widehat\beta)}n.}
$$
If the fitted residual sum is zero, the likelihood instead increases without bound as $\sigma\downarrow0$: there is no joint maximum with the stated positive scale parameter. The displayed estimate then denotes the boundary limit, not an admissible positive-scale MLE.
An ordinary <exponential dispersion family> has log density $[y\theta-b(\theta)]/a(\sigma)+c(y,\sigma)$, so changing the mean changes the log density by an affine function of $y$. For two distinct Laplace locations $\mu_1,\mu_2$, their log-density difference is $(|y-\mu_2|-|y-\mu_1|)/\sigma$, with different slopes on the intervals separated by the two locations. It is not affine in $y$. Equivalently the kink at $y=\mu$ moves with the parameter. \b[The Laplace location family is not an ordinary <exponential dispersion family>.]
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