If $X$ and $Y$ are two independent random variables and $X$ and $X-Y$ are also independent then $X$ is a constant almost surely. I have been thinking about this problem but I haven't found an answer.

Problem. I have two random variables $X$ and $Y$ which are independent  ($X \perp Y $)  in $( \Omega, \mathcal{F}, P  )$. In addition to that, $X$ and $X-Y$ are also independent ($X \perp X-Y $) .
I have to prove that $X$ is a constant almost surely ($\exists c \in \mathbb{R} : P(X=c) = 1$). That is, $X$ is a degenerate random variable.

I suppose I should take the supremum of a set. Maybe I should try to discover that $P(X\leq0)=1$ or $P(X\leq0)=1$.
Would you mind giving me some ideas or hints?
 A: I will use characteristic functions to avoid use of moments.
Hints: We have $Ee^{itY}=Ee^{it[X-(X-Y)]}=Ee^{itX} Ee^{-it(X-Y)}=Ee^{itX}Ee^{-itX}Ee^{itY}$. Since $Ee^{itY}\neq 0$ for $|t|$ sufficiently small, this gives $|Ee^{itX}|^{2}=1$   for $|t|$ sufficiently small. This implies that $X$ is almost surely constant.  Ref. Corollary on p. 75 of A Course in Probability Theory by K L Chung.
A: For any bounded continuous $\varphi(\cdot)$, define $\widetilde{\varphi}(x) = \mathbf{E}[\varphi(x - Y)]$. Also, let $B$ be any Borel subset of $\mathbb{R}$. Then by the independence of $X-Y$ and $Y$,
\begin{align*}
\mathbf{E}[\varphi(X-Y) \cdot \mathbf{1}_{\{X \in B\}}]
&= \mathbf{E}[\varphi(X-Y)] \cdot \mathbf{E}[\mathbf{1}_{\{X \in B\}}] \\
&= \mathbf{E}[\mathbf{E}[\varphi(X-Y) ] \cdot \mathbf{1}_{\{X \in B\}}]
\end{align*}
On the other hand, by conditioning on $X$ and utilizing the independence of $X$ and $Y$,
\begin{align*}
\mathbf{E}[\varphi(X-Y) \cdot \mathbf{1}_{\{X \in B\}}]
&= \mathbf{E}[ \mathbf{E}[ \varphi(X-Y) \cdot \mathbf{1}_{\{X \in B\}} \mid X]] \\
&= \mathbf{E}[\widetilde{\varphi}(X) \cdot \mathbf{1}_{\{X \in B\}} ].
\end{align*}
Since this is true for any $B$, by comparing both sides, we get
$$ \mathbf{E}[\varphi(X-Y) ] = \widetilde{\varphi}(x) = \mathbf{E}[\varphi(x - Y)] $$
for $\mathbf{P}(X\in\cdot)$-almost every $x$. Since $\varphi$ is arbitrary, this then implies that
$$X-Y \stackrel{\text{d}}= x - Y $$
for $\mathbf{P}(X\in\cdot)$-almost every $x$. However, this can only make sense when $X$ is degenerate, for otherwise we would have been able to find $x \neq x'$ such that $x - Y$ and $x' - Y$ have the same distribution. Therefore the claim follows.
A: $\text{cov}(X,X-Y) = \text{var}(X) - \text{cov}(X,Y)$
If $X,Y$ are independent $\text{cov}(X,Y) = 0$
Which implies $\text{var}(X) =0$
