# Congruence Relations that relate to Multinomial Theorem

Defining the multinomial expansion as follows:

$$\left( \sum _{j=1}^m x_j \right)^n = \sum _{j=1}^{T_{n,m}} \left( C_{n,m,j} \prod _{i=1}^m {x_i}^{\alpha_{n,m,i}} \right) \tag 0$$

The following two lemmas:

$$n \left( {\frac {\sum_{j=1}^{T_{n,m}} C_{{n,m,j}}}{n-1}}-m \left\lfloor {\frac {\sum _{j=1}^{T_{n,m}}C_{n,m,j}}{ \left( n-1 \right) m^2 n}}\right\rfloor \right)=1 \tag 1$$

$$n-m \Biggl\lfloor \frac{n}{m} \Biggr\rfloor =1 \Longleftrightarrow n^k\equiv 1 \pmod m \quad \forall k \in \mathbb N \land n,m \in \mathbb N \setminus \{1\} \tag 2$$

were used to establish:

$$\left( \frac {\sum^{T_{n,m}}_{j=1}C_{n,m,j}}{(n-1) m} \right)^k \equiv 1 \pmod m \tag{A1}$$

$$\left( \frac {\sum_{j=1}^{T_{n,m}} C_{n,m,j}-m^n}{n-1} \right)^k \equiv 0 \pmod m \tag{A2}$$

And I am also seeking similar routes to proof for:

$$\sum _{j=1}^{T_{n,m}}C_{n,m,j} \equiv m^n \pmod m \tag{B1}$$

$$T_{n,m}n\equiv0 \pmod m \tag{B2}$$

Edit: This identity will suffice for proof of the above congruence:

$$T_{n,m}=\frac{(n+m-1)!}{(m-1)!n!} \tag{B2*}$$

$$\sum ^{T_{n,m}}_{j=1}C_{{n,m,j}}\frac{(n+m-1)!}{m!(n-1)!}\equiv0 \pmod m \tag{B3}$$

• How did you write your MathJax code? I understand that there are software packages somewhere that do things like that for you, although I have no idea how people who post here find out about those nor what any of them are called, and the loony way your code was written looks like the output from some of those, but there were also things in it that looked like someone would do manually if they're not particularly skilled in MathJax or LaTeX coding (e.g. the way in which you wrote the equation numbers and the left-right-arrow. $\qquad$ – Michael Hardy Nov 14 '19 at 16:47
• Sorry so what is the "skilful" manner on rights equation labels? – Adam L Nov 14 '19 at 16:52
• Look at my edits to the question. – Michael Hardy Nov 14 '19 at 16:53
• righto I guess I will remember to do it that way now that I know – Adam L Nov 14 '19 at 16:54