Let $p, q$ be distinct primes and $L := \mathbb{Q}(\sqrt[3]{p}, \sqrt{q})$.  I want to show that $L = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})$. Of course one inclusion is clear, but how do I get the other one?
121085
Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$  So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.  This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.  Could someone maybe give me a hint about how to proceed? Thanks a lot!!!
101930
I have this algorithm and Im searching p and q:  $n=p^2 * q $  $l=(p-1)*(q-1) / \gcd (p-1,q-1)$  $d\equiv l^{-1} \pmod n$  And the values are:  For $n$:  1043247381320041410146374847683905608886067988802815759655490071506964811427733496668055364700822779908166748193534341903285387332658081562041534228961538451830025041762537639985150344917584064827615356535763872539925989984117068207555498700585376330731428359489791600319636079777025344217248475485835298689437926003958578141784379857951264973109214803916445702964047899447002191548965239884095802091798787988745415178578597206815122915438351636672681355517982410123569961242563194480399347961753678159210217993526738306796503660563163029287419732210247579793266891085178817047536758465239950861266592524206390869822972469137566045176690277813352563392838847417674529251627419479815097868890993153215515488877097760272399483111013602618554846096893805108827211344467797072352072430591318485091677011261912992157762106672216780766138057271434398274272110062768324672899209610001835423246735718435995420720133657793744810610657   for $d$:  105717877672224295784291675385546750129601472163379527375367128238003471793097341763391999503913659542087520808364287167645349555897774428176288829031477703594776341268160433668976237198330124904372546897953955241778475322074160469767162566234358907740412805211247598698481640207956827199048419949848619731050860757808082106844953602176503247786535423423104239566910713528691298692332149783689032722772009974404716783312200885555343877611094998117307970399241640213553820043460140493803404190774171912962651075100983834071259021038322886114660028788229094872545516054757700068264628024524394995795135464746570071201   How can I get $p$ and $q$?   Thanks!
109318
We know that for $m=p^r, 1-\zeta_m$ is a prime.Now suppose that m has at least 2 distinct primes appearing in its prime factorization,we need to show that $1-\zeta_m$ is a unit in its ring of integers $\mathcal{O}_{\mathbb{Q}(\zeta_m)}=\mathbb{Z}[\zeta_m]$.  I tried proving that $N_{\mathbb{Q}}^{\mathbb{Q}(\zeta_m)} (1-\zeta_m)=\pm 1$ but got stuck in finding norm of $\zeta_m$. Some hint would be nice.
79189
Prove that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes.   I am a self taught person. I just learned this and tried this on my own and came up with this.  $x \equiv 1 \pmod{p}$ and $x \equiv -1 \pmod{q}$ has a solution $[a]_{pq}$, since $p$ and $q$ are relatively prime. Because $q$ is an odd prime, $[-1]_{pq}$ is not a solution, so $[a]_{pq}\neq [-1]_{pq}$. But $a^2 \equiv 1 \pmod{p}$ and $a^2 \equiv 1 \pmod{q}$, so $a^2 \equiv 1\pmod{pq}$ since $p$ and $q$ are relatively prime, and thus $[a]_{pq}$ has order 2.  Can someone please tell me if this proof this correct? Please help with proof as I learned it just now as a self taught person.
841
I don't know about the Sylow Theorems.  But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.  If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.   But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.  I am asking for help completing this argument.  Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).
86086
I want to show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$.  I think it would be easier to prove it using the following: $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{3})$. Then $\sqrt{2}\notin\mathbb{Q}(\sqrt{3})$ so $\sqrt{2}\in \mathbb{Q}(\sqrt[4]{3})\Longleftrightarrow \sqrt{2}=a+b\sqrt[4]{3}$, with $a,b\in\mathbb{Q}(\sqrt{3})$.   I tried simplifying the above equation, but I did not get anything.
101932
I'd really love your help with proving that for  $q, p$ odd primes such that $p \neq q$, there is not primitive root modulo $pq$.   There's a theorem says that there's primitive root only for $2, 4, q^l, 2q^l$ for prime $q$ and $l \in {Z}$, but I guess I can't use it.  any other suggestions?  Thanks a lot
59795
$p,q$  primes, $(pq,a)=1$ and $p-1|q-1$ then prove $a^{q-1}\equiv 1\pmod{pq}$   Got this in an exam today and had no idea how to do it.
89939
The question is:   Find two distinct odd primes p and q such that  $({p-1})^{q-1}+({q-1})^{p-1} \equiv 101 \pmod {pq}$ Well... I only have the idea to write down     ${p-1}^{q-1} \equiv -1^{q-1} \equiv 1 \pmod {p}$   Actually, I solve this question by brute force and get p=11 q=3; however, I wanna know some wise solutions...  Mine is .... a bad one  Please help me to solve this question :(
99288
Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$   $(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$   $(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$  $(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated. Thanks for your time.
599
This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts).  Here's the full problem:  Let $x_{1}, x_{2}, x_{3}, \cdots x_{y}$ be all the primes that are less than a given $n&gt;1,n \in \mathbb{N}$.   Prove that $$x_{1}x_{2}x_{3}\cdots x_{y} &lt; 4^n$$  Any ideas very much appreciated!
84593
Suppose we have a set of primes $p_1,\dots,p_t$. Prove that  $\log p_1,\dots,\log p_t$ is linear independent over $\mathbb{Q}$. Now, this implies $ \sum_{j=1}^{t}x_j\log(p_j)=0 \iff x_1=\dots=x_t=0$.  I think I have to use that fact that every $q\in\mathbb{Q}$ can be written as $\prod_{\mathcal{P}}$, where $n_p$ is a unique sequence ($n_2$,$n_3$,$\dots$) with domain $\mathbb{Z}$. Here, $\mathcal{P}$ denotes the set of all integers.  Now how can I use this to prove the linear independency?
45734
$p_1, p_2$ are distinct prime numbers  I have just observed this pattern when solving  problem. Is there a simple way to prove/disprove it ?
105272
It seems like $$\log n \leq \sqrt n \quad \forall n \in \mathbb{N} .$$ I've tried to prove this by induction where I use $$ \log p + \log q \leq \sqrt p \sqrt q $$ when $n=pq$, but this fails for prime numbers. Does anyone know a proof?
51735
I'm having difficult proving this.  As a hint the exercise to prove first, that if $a\lneqq \pm 1$ satisfies $a \equiv 3 \pmod4$, then exist $p$ prime, $p \equiv 3 \pmod 4$ such $p\mid4$. But I'm not really getting for what purpose can this be used.
120637
Here's a question I've been asked;  Given distinct primes $p,q,r$, show that any group $G$ of order $pqr$ is not simple.   So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only $$\dbinom{3}{2} + \dbinom{3}{1} + 1$$ of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of $n$ distinct primes.  Is there some other approach? Thanks for any insights
78124
Assume that $p$ and $q$ are distinct add primes such that $p-1\mid q-1$. If $\gcd(a,pq)=1$ ,show that: $$a^{q-1} \equiv 1 \pmod{pq}$$   I have tried as follows: $$a^{q-1} \equiv 1 \pmod{q} \quad \text{and} \quad a^{p-1} \equiv 1 \pmod{p}$$ $$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{q} \quad \text{and} \quad a^{(q-1)(p-1)} \equiv 1 \pmod{p}$$ $$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{pq}$$ But then I am stuck - please help.