Prompt. Let $X_1, X_2, \dots, X_n$ be an i.i.d. sample from the Pareto distribution (i.e. $f(x; a, b) = \frac{1}{b} e^{-\left(\frac{x-a}{b}\right)} I_{(a, \infty)}$).
Solution. We'll first analyze $\dfrac{f(x;a,b)}{f(y;a,b)}$ in order to determine independence in relation to $T(\mathbf{x}) = T(\mathbf{y})$.
$$\frac{f(x;a,b)}{f(y;a,b)} = \dfrac{\frac{1}{b} e^-(\frac{x-a}{b})}{\frac{1}{b} e^-(\frac{y-a}{b})} = \frac{e^\frac{y-a}{b}}{e^\frac{x-a}{b}}$$
This ratio is constant when $y = x$ and consequently when $\sum\limits_{i=1}^n x_i = \sum\limits_{i=1}^n y_i$. Thus $\sum\limits_{i=1}^n x_i$ is a minimum sufficient statistic.
Question. Is this correct? I feel like I'm missing something, since I usually struggle with the problem sets.
HINT
You are missing something. You forgot to include the $I(a,\infty)(x)$ factor. It depends on $x$ and $a$ so it probably matters for inference for $a.$
You can write the likelihood as $$ L(a,b;x_1,\ldots, x_n) = \dfrac{e^{na/b}}{b^n} e^{-\frac{1}{b}\sum_i x_i } I(a<x_1)I(a< x_2)\ldots I(a < x_n)$$ so $\sum_i x_i$ is sufficient for $b.$ Now can you simplify the $I$ factors to find a statistic sufficient for $a$?