Let $X_1,X_2,X_3$ be independent such that for all $x > 0$,
$$ P(|X_i| > x) < e^{-x}, \;\;\; i = 1,2,3 $$
Prove that if $X_1+X_3$ and $X_2+X_3$ have the same distribution, then so does $X_1$ and $X_2$.
Because of the condition given, I assumed the question meant for us to use moment generating functions. Since $X_1+X_3$ and $X_2+X_3$ have the same distribution, their MGFs are equal for all $t$,
$$ \phi_{X_1}(t)\phi_{X_3}(t) = \phi_{X_1+X_3}(t) = \phi_{X_2+X_3}(t) = \phi_{X_2}(t)\phi_{X_3}(t)$$
where the first and last equalities come from the independence assumptions. EDIT: If I can show that $\phi_{X_3}(t)$ is finite in a neighbourhood of $0$, say $(-\delta,\delta)$, then $\phi_{X_1}(t) = \phi_{X_2}(t)$ for all $t \in (-\delta,\delta)$. By the uniqueness theorem, this would imply that the distributions of $X_1$ and $X_2$ are the same.
EDIT: To prove finiteness, first suppose $t > 0$. Then,
$$ E[e^{tX_3}] = \int_0^\infty P(e^{tX_3}>x)dx = \int_0^\infty P\left(X_3 > \frac{1}{t}\ln(x)\right)dx \leq 1 + \int_1^\infty x^{-\frac{1}{t}}dx $$
The above is finite when $t<1$. We can do something similar when $t < 0$.