= Solution
Expanding the exponent gives
$$
f_\mu(x_1,x_2)
=\frac1{2\pi}e^{-(x_1^2+x_2^2)/2}
e^{\mu(x_1+x_2)-\mu^2}.
$$
The <Fisher-Neyman factorization theorem> therefore shows that $T=X_1+X_2$ is sufficient.
It is also minimal sufficient. For two sample points $x,y\in\mathbb R^2$,
$$
\frac{f_\mu(x)}{f_\mu(y)}
=C(x,y)\exp\left(\mu[T(x)-T(y)]\right),
$$
where $C$ is independent of $\mu$. This ratio is independent of $\mu$ exactly when $T(x)=T(y)$. The <likelihood-ratio criterion for minimal sufficiency> applies, proving the claim and the general <normal sample sum with known variance> result in this case.
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