So I am answering a problem where $\underline{X}=(X_1,X_2)'$ be a random variable from $N(\mu,1).$
Here is my approach, the joint density is:
$f(x_1,x_2|\mu) = (2\pi)^{-1} \textrm{exp}(\frac{-1}{2\sigma^2}\sum^{2}_{i=1} x^2_i + \frac{\mu}{\sigma^2} \sum^2_{i=1}x_i-\frac{2\mu^2}{2\sigma^2})$
Since the variance is known as a constant 1 then $$h(\textrm{x}) = (2\pi)^{-1} \exp(\frac{-1}{2})\sum^2_{i=1} x^2_i$$ and $$g_u(\bar x,n) = \exp(-\mu^2 + \mu r(x_1.x_2))$$ where $r(x_1,x_2) =\sum^2_{i=1}x_i$
Therefore it can be concluded that sample mean $\bar X_n $ is also a sufficient statistic.
But then after concluding it, the problem asks if $S_2 = S_2(\underline{X})= X_2 - X_1$ is sufficient for $\mu$. Should I still need to prove it ? Or nay nay ?
$$S_2=X_2-X_1\sim N(0;2)$$
Its distribution is is the same $\forall \mu$ thus $S_2$ is ancillary for $\mu$.
In other words, by itself it can tell us nothing about $\mu$
...it is not sufficient
Also note that $S_2=X_2-X_1$ is independent from $S_1=X_1+X_2$.
This because
$S_1$ is Complete and Sufficient for $\mu$
$S_2$ is Ancillary for $\mu$
Thus we can apply Basu's Theorem because all the hypothesis are attained