Proof that $$ \begin{bmatrix} 3 \\p\end{bmatrix} = \begin{cases} 1 & p \equiv1,11\mod 12 \\ -1 & p \equiv5,7\mod 12\end{cases}$$
Consider the group $G$. An involution in $G$ is an element $a \in G$ of order two, i.e. $a^2 = e$, with $e$ the identity in $G$. Suppose that all elements ($e$ excluded of course) of $G$ are involutions. Proof that $G$ is abelian. Can you say something more about $G$ with the assumption that $G$ is finite?
A palindromic number is a number that remains the same when its digits are reversed. Proof that all palindromic numbers with an even number of digits are always divisible by 11. Is the converse also true (proof/counterexample).
Suppose that $X$ is a set with cardinality $n$ and that $U$ is a set of subsets of $X$ with cardinality $d$ $(d\le n)$. The elements of $U$ are chosen in a special way so that all subsets with cardinality $t$ ($t$ is a fixed number, $d \le t \le n$) contain exactly $1$ element of $U$. Suppose that $S$ is a subset of $X$ with $|S| = i, t\le i \le n$. Find the number of elements of $U$ that are contained in $S$.
I'm struggling with answering these questions. Any hints? It doesn't matter if you don't know all of the answers to each question. All help is welcome!
An enormous 'thank you'!
EDIT (here are my attempts, by popular demand):
1) We want to prove that $x^2 \equiv 3$ has a solution $\operatorname{mod} p$ if $p \equiv 1, 11, \mod 12$. Is this proof similar to proving -1 is a square modulo $p$? But where do the specific numbers 1, 11, 5 and 7 come from? Why modulo 12?
2) Take $g,h \in G$ randomly. We want to prove that $gh=hg$. Since $g$ and $h$ are elements of $G$, these elements have an inverse $g^{-1}$ and $h^{-1}$ resp. These inverses are also involutions. $G$ is a group, so the product of elements in $G$ is also an element in $G$. We get: $(gh)(gh)=e = (gh)(h^{-1} g^{-1})$ and $(hg)(hg)=e=(hg)(g^{-1} h^{-1}) $. This means that $(gh)^2 = (hg)^2$. Is this enough to conclude that $gh=hg$? Do I actually need the inverses, because I haven't really used them?
3) I have no clue here. Maybe something like this: suppose we have a palindromic number $n$ with $2k$ digits. When $k=1$, $n/11$ will give as a quotient consisting of one number from 1 to 9. For $k=2$, we'll have a quotient of 2 digits etc. I guess it's also useful to consider the set of residue classes modulo 11.
4) Honestly, I don't even understand the question.