Modular algebra problems

Question

I got some problem with those demonstrations and I don't know where I'm wrong, let me show you my steps:

1: first of all

$ 6 | 2n(n^2 +2) $

That is, I must demonstrate that $6$ divides $2n(n^2 +2)$ , so I just simplify this stuff dividing by two:

$ 3 | n(n^2+2) $

to solve this I thought to model this with a simple modular algebra problem of the type:

$ n(n^2+2)≡0 ($mod $ 3)$

That is, I must try to satisfy this system of modular equations:

$ n≡0($mod $ 3)$

$ n^2≡-2($mod $ 3)$

but the amount of numbers that are smaller than 3 and co-prime with it are 2, and $gcd(2,2)=2$ meaning that we can't find the inverse of the application $x → x^2$.

2: Let $A:=Z/50Z$ and let $B:= (Z/50Z)*$ be the subset of A made of its invertible classes (mod 50). Calculate the cardinality of the following sets:

$X:=C∈2^A|B⊂C$;

$ Y:=f∈A^A|f(B) =B $;

$ Z:=f∈Y|$ f is injective.

with this I don't even know where to start :(

Answer 1

Hint:

1. Use induction

2. $x\in B$ iff $(x,50)=1$ (coprime). To count the number, you need the Euler Phi function.

$X$ is the collection of all subsets of $A$ that contains $B$. To define one such subset, you need to assign $0$ or $1$ to each element outside $B$ ($1$ means the subset has it and $0$ means it does not), and therefore $|X|=2^{|A|-|B|}$.

$Y$ is the collection of all functions from $A$ to $A$ that fixes $B$. Basically, you first need to assign value $f(x)$ for each element $x\not\in B$, and the number of choices is $|A|^{|A|-|B|}$. Next, $f|B$ is surjective and therefore also bijective, and therefore $\{f|B: f(B)=B\}$ is the permutation group of $B$, with cardinality $|B|!$. Overall, you have $|Y|=|A|^{|A|-|B|}\cdot |B|!$.

$Z$ is injective functions from $A$ to $A$. You need to ensure $f|B$ and $f|(A-B)$ are both injective, and therefore both bijective. Therefore, $|Z|=|B|!\cdot (|A|-|B|)!$.

Answer 2

You may try below for first problem :

$n(n^2+2) = n(n^2-1+3) = n(n^2-1) + 3n = (n-1)n(n+1) + 3n$