on counting certain number of 4-tuples Let $i,j,k,l \in \{0,1,2,3,\dots,n\}$. How can we count the number of $4$-tuples $(i,j,k,l)$ with the property $i-j=k-l$ ? Thanks for any hints/responces

$i+l=:t=j+k$
So fix $t$. This means that you have to take $(i,l)$ and $(j,k)$ such that their sum is equal to $t$.
Thus we have exactly $t+1$ possibilities for $(i,l)$ and $t+1$  for $(j,k)$. Hence the number of possibilities is equal to $(t+1)^2$ for $(i,j,k,l)$. Here $t$ will go from $0$ to $2n$ and so the total number $N$ of possibilities for $(i,j,k,l)$ such that $i-j=k-l$ is equal to
$$N=\sum_{t=0}^{2n} (t+1)^2=\sum_{k=1}^{2n+1}k^2=\frac{(2n+1)(n+1)(4n+3)}{3}$$
Here I’ve used te closed form formula of the sum of squares
$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$
