How many Lorentz scalars are needed to characterise $n$ 4-vectors? If I have an arbitrary function of $n$ 4-vectors $f = f(q_1^\mu, q_2^\mu, ..., q_n^\mu)$ where $q_i^\mu$ are 4-vectors, what is the least number of Lorentz scalars I would need if I needed to specify the function's argument in the form $f = f(s_1, s_2, ..., s_k)$ where $s_i$ are Lorentz scalars? These k scalars could then be said to characterise the n 4-vectors and we could construct any arbitrary scalar function $f(q_1^\mu, q_2^\mu, ..., q_n^\mu)$ from these.
For context, this question arose from a book by deGroot which takes a function $W(p_1, p_2, p_1', p_2')$ which is a Lorentz scalar and then says that this function can be constructed out of 10 scalar invariants. I can see that at least the 8 products $p_i^\mu p_{j\mu}$ could be such scalar invariants but I can't think of anything more. If you can give a general argument for n 4-vectors then it would help me understand how these extra scalars could be constructed. 
 A: There are 6 ways to pick two things out of 4, and 4 ways to pick one thing out of 4. So 10 altogether:
p1 p2
p1 p1'
p1 p2'
p2 p1'
p2 p2'
p1' p2'
p1 p1
p2 p2
p1' p1'
p2' p2'
Hence the number of easily available scalar invariants here is 10 not 8 as you guessed.
This is just me making up what seems to me to be a reasonable answer. If it is wrong then I would be happy to correct it; I would learn something myself.
A: $$
\frac12 n (n + 1)
$$
The easiest way to see it is to consider the number of independent elements of the $n\times n$ symmetric matrix $m_{ij} \equiv p^\mu_i p_\mu^j$. The entries of the matrix clearly exhaust all possible Lorentz contractions.
The number of independent elements is $1 + 2 + \cdots + n = \frac12 n (n + 1)$, which can easily be seen by counting the numbers of diagonal elements starting from e.g. the top right, and stopping once we reach the main diagonal.
A: The scalars can be constructed by taking products of the 4-vectors taken two at a time. So $p_i^\mu p_{j \mu}$ constitute all the Lorentz scalars possible.
The number of these scalars are therefore $^nC_2 + ^nC_1$ represent all the possibilities since


*

*There are $^nC_2$ ways to make a product $p_i^\mu p_{j \mu}$ where $i \neq j$.

*There are $^nC_1$ ways to make a product $p_i^\mu p_{i \mu}$.


As to why these are the only scalars possible, the argument is that each of these vectors comprises of a magnitude and a direction in 4-space. Their magnitudes are given by their products with themselves as in (2). Their directions must be seen relative to each other. This sense of relative direction is imbued in their inner products with each other which is counted by (1).
