Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$ 
Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$.  Find the joint PMF of $X, Y, N$.

I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.
The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = \binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.
I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write
\begin{align}
P(X=x, Y=y, N=n) &= P(X=x, Y=y \mid N=n)P(N=n) \\
&= P(X=x, Y=y \mid N=x+y)P(N=n),
\end{align}
because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,
\begin{align}
P(X=x, Y=y, N=n) &= P(X=x \mid N=x+y)P(N=n)
\end{align}
I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y \mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $\frac{1}{x+y+1}$. Hence,
\begin{align}
P(X=x, Y=y, N=n) &= \frac{1}{x+y+1} (n+1)(1-p)^n p^r \\
&= (1-p)^n p^2.
\end{align}
But this looks like nonsense.

We have,
$$
P_{X,Y,N}(x,y,n) = 
\begin{cases}
P(X=x).P(Y=y) & if x+y=n \\
0 & otherwise
\end{cases}
$$
So,
\begin{align*}
    P_{X,Y,N}(x,y,n) = P(X=x, Y=y, N=n) 
                     = P(X=x).P(Y=y)
                     = pq^x.pq^y
                     = p^2q^{x+y}
                     = p^2q^{n}
\end{align*}
Hence, the joint PDF of $X$, $Y$ and $N$ is
$$
P_{X,Y,N}(x,y,n) = 
\begin{cases}
p^2q^{n} & if x+y=n \\
0 & otherwise
\end{cases}
$$
