Prove that $(\mathbb{Z}/23\mathbb{Z})^*$ is generated by $5\pmod{23}$   Is there a tactic or theorem that could help me finding the power of $5\pmod{23}$ for each element?  I was thinking about using Euler's theorem or Fermat's Little theorem, but I don't see how they could help me get for example $4\pmod{23}$ as generated by $5\pmod{23}$.   Thanks in advance
124235
I'm stuck on that question, because I'm trying to find a simple way to solve it.  I'm pretty sure there exists such elements because the group is cyclic and $|\mathbb{Z}_{31}^{*}| = 30$, hence there exists a sub-group of any order dividing $30$. So I know there is a sub-group of order $10$ and it's cyclic because any sub-group of a cyclic group is cyclic, and it's generator is an elements of order $10$.      Now, how can I proceed from here? How can one find the generator of this sub-group?
70953
I believe that this may require the use of Fermat's Little Theorem. I rewrote it as $(a+1)^p - a^p \equiv 1 \pmod{p}$ because the right-hand side looks similar to Fermat's Little Theorem, but I was unable to figure out how I can get the left-hand side to become $a^{(p-1)}$.
114214
How to show that there is an element of order $51$ in the multiplicative group $(Z/103Z)^∗.$  I can't use Cauchy's theorem since $51$ is not prime.
25328
Show that the set $\{5,15,25,35\}$ is a group under multiplication modulo $40$. What is the identity element of this group. Can you see any relationship between this group and $\mathbb U(8)$?   I am very stuck on this question and I think my knowledge of abstract algebra is my liability at the moment. Right now I'm most confused by the fact that I learned: An element of a group of this nature must be relatively prime with the working mod.  Obviously here this is not the case, so I don't really know where to begin.  Do I need a Cayley Table?  Thanks.
78572
How do i prove that $17^n−12^n−24^n+19^n≡0(\mod35)$ for every possitive integer n. Can anyone give me a hint of how to start?
86566
Fix $k \in \mathbb{Z}_+$. Prove that we can find infinitely many primes of the form $a2^k +1,$ where $a$ is a positive integer. We can use the result that: If $p \ne 2$ is a prime, and if $p$ divides $s^{2^t}+1$, for $s &gt; 1$ and $t \ge 1$, then $p \equiv 1 \pmod {2^{t+1}}$.  Ive trying to get something inductively: For $k = 1$, there are infinitely many primes of the form $2a + 1$. Suppose there are infinitely many primes of the form $a2^k + 1$, and then show that there are infinitely many primes of the form $a2^{k+1} + 1$. If there are infinitely primes of the form $a2^k + 1$, where $a$ is even, then we have that, $a2^k + 1 = (2q)2^k + 1 = q2^{k+1} + 1$. Hence we are done. Therefore, suppose there are only infinitely many primes of form $a2^k + 1$, where $a$ is odd. - but I can't get a contradiction out of this.
74113
Let $n&gt;1$ be an integer. Prove that $0$ is a radical ideal in $\mathbb{Z}/n\mathbb{Z}$ if and only if $n$ is a product of distinct primes to the ﬁrst power (i.e., $n$ is square free). Deduce that $(n)$ is a radical of $\mathbb{Z}$ if and only if $n$ is a product of distinct primes in $\mathbb{Z}$.  I don't really get it. For example, $8$ is not a square-free integer; so take $\mathbb{Z}/8\mathbb{Z}$. Why is $0$ not a radical ideal in this ring but is in $\mathbb{Z}/6\mathbb{Z}$?
1549
I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$  But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
75989
Here is the question I want to answer: Let $R = \mathbb Z[\sqrt{-5}]$ Show that $I = (2, 1 + \sqrt{-5})$ is not a free $R-$module. Hint: It helps to view $R$ as $\mathbb Z[x]/(x^2 + 5).$ My questions are:: 1- How can I use the hint? I have found here  how to show that it is not free as it says in the answer in that link the following: It is generated by two elements so if it were free, it would be free of rank $1$ or $2$. If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$. If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent. 2- for showing this &quot;If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.&quot;, should I use the norm function? Could someone push me in the right direction to answer those questions please?
63022
Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$  So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.  Any hint would be greatly appreciated!
7687
I came across the following in the book Handbook of Mathematical Induction: $$ 19\mid (2^{2^n}+3^{2^n}+5^{2^n}),\quad n\in\mathbb{Z^+}\tag{1} $$ Apparently, this problem is not so bad if you think about it in a somewhat "outside of the box" sort of way, but it is supposed to be quite challenging otherwise.   Question: How might one prove $(1)$ using induction or otherwise? The "outside of the box" answer has been provided but with details oranged out so as not to spoil it for anyone who may come up with that solution individually.
104600
Show that $x^{20}=1$ for all $x\in U(100)$.  In the book that I'm reading, it says "Since $U(100)\thickapprox Z_2\oplus\,Z_{20}$ we see that $x^{20}=1$ for all $x$ in $U(100)$".I think maybe that remark is due to this theorem which states "The order of an element of a direct product of a finite groups is the least common multiple of the orders of the components of the elements". i.e $$|(g_1,g_2,.\,.\,.,g_n)|=lcm(|g_1|,|g_2|,.\,.\,.,|g_n|).$$ We define $U(n)$ to be the set of all positive integers less than $n$ and relatively prime to $n$ with the operation  multiplication modulo $n$
28336
Find all positive integers $n$ such that $a \equiv a^{-1} \pmod{n}$ for all invertible $a$ modulo $n$.   I found that $n = 1,2,4,6,8,12,24$ satisfy this. How can we prove that these are all of them?
76548
If $a^n\equiv a \pmod n$ for all integers $a$, does this imply that $n$ is prime?  I believe this is the converse of Fermat's little theorem.
66574
Show that $13$ divides $2^{70} + 3^{70}$.   My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?  Thanks!
4812
Problem: $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$  My work: Start by letting $m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$. If $(a,p_i)=1$ for some integer $i$, then we have that  $$ a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} $$ by Euler's theorem. Since $\phi(p_i^{a_i}) \mid \phi(m)$, we have that $p_i^{a_i}\mid (a^{\phi(m)}-1)$ if $(a,p_i)=1$.   Beyond this, I am stuck at the moment.
27206
a, b are integers. p is prime. I want to prove: $(a+b)^{p} \equiv a^p + b^p \pmod p$  I know about Fermat's little theorem, but I still can't get it I know this is valid: $(a+b)^{p} \equiv a+b \pmod p$ but from there I don't know what to do.  Also I thought about $(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1} \binom{p}{k} a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$ Any ideas?  Thanks!