Here mean and variance are tied together, both equal to $\theta$ , so the model has a single parameter. Expand the exponent of one observation:

$-\frac{(x-\theta)^{2}}{2\theta} =-\frac{x^{2}-2x\theta+\theta^{2}}{2\theta} =-\frac{x^{2}}{2\theta}+x-\frac{\theta}{2}.$

The middle term $x$ carries no $\theta$ . Hence the joint density is

$L(\theta)=(2\pi\theta)^{-n/2} \exp\Bigl\{-\frac{1}{2\theta}\sum_{i=1}^{n}x_i^{2}\Bigr\}\cdot \underbrace{\exp\Bigl\{\sum_{i=1}^{n}x_i\Bigr\}}_{\text{free of }\theta} \cdot e^{-n\theta/2}.$

By the factorisation theorem, with $h(\uline{x})=\exp\{\sum x_i\}$ and $g\bigl(T;\theta\bigr)$ depending on the data only through

$T=\sum_{i=1}^{n}X_i^{2},$

the statistic $T$ is sufficient. This is a full-rank one-parameter exponential family with natural parameter $-1/(2\theta)$ and natural statistic $\sum x_i^{2}$ , so $T$ is also minimal sufficient.

Why the others fail: $\overline{X}$ alone is not sufficient (the likelihood cannot be reconstructed from it), and the pair in (C), while sufficient, is not minimal — it is a function-of-a-function coarser than needed, since $\sum X_i^{2}$ alone already suffices. The lesson is that constraining $\sigma^{2}=\mu$ collapses the usual two-dimensional sufficient statistic down to one dimension.