There is a  however that question asks why $3 |p^2$. Here the question is about $ 3 | p^2 \rightarrow 3 | p$.   It is a simple exercise (1.2.1) from Abbot's "Understanding Analysis".   $\nexists p,q \in \mathbb{N} : \left(\dfrac{p}{q}\right)^2 = 3$   $p$ and $q$ have no common factors, otherwise they would cancel each other out.  Contradiction:  $p^2 = 3q^2$   The troublesome part for me is      From this, we can see that $p^2$ is a multiple of 3 and hence $p$ must   also be a multiple of 3.   Why is that?  The proof goes on with   $p = 3 r$  $q^2 = 3r^2 \rightarrow q = 3 \lambda$.   It concluses now that $q$ and $p$ have common factors, which invalidates the original statement.   How can I prove $p^2 = 3q^2 \rightarrow p = 3 r$? If this is not true, then I don't see how the contradiction proves the statement, since it only invalidates it for such $p,q$ that have common factors.   What if $p^2 = 3q^2$ and $p \ne 3 r$?
121963
Assume that  $$3 = \frac{p^2}{q^2}$$  So,   $$ 3 q^2 = p^2$$  So $p^2$ is divisible by $3$. How we can conclude this?
29040
The question given to me is:     Show that $\large\frac{(3^{77}-1)}{2}$ is odd and composite.   We can show that $\forall n\in\mathbb{N}$:  $$3^{n}\equiv\left\{    \begin{array}{l l}     1 &amp; \quad \text{if $n\equiv0\pmod{2}$ }\\     3 &amp; \quad \text{if $n\equiv1\pmod{2}$}\\   \end{array} \right\} \pmod{4}$$  Therefore, we can show that $3^{77}\equiv3\pmod{4}$. Thus, we can determine that $(3^{77}-1)\equiv2\pmod{4}$. Thus, we can show that $\frac{(3^{77}-1)}{2}$ is odd as:  $$\frac{(3^{77}-1)}{2}\equiv\pm1\pmod{4}$$  However, I am unsure how to show that this number is composite. The book I am reading simply states two of the factors, $\frac{(3^{11}-1)}{2}$ and $\frac{(3^{7}-1)}{2}$, but I do not know how the authors discovered these factors.   I'd appreciate any help pointing me in the right direction, thanks.
119233
I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$    But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement.  My question: What is wrong with my argument?  And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.
13378
More specifically, I have to find a $q$ in $A$ such that $q$ is larger than any $p$ in $A$.  The only thing I can think of is using (${2 - {p^2}}$) somehow, or ${(p+x)^2}&lt;2$, but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise:     To do this, we associate with each rational $p &gt; 0$ the number:     $$q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}$$   But he didn't explain how he came up with that number.   The only thing I've thought of is the following procedure:  $$\eqalign{   &amp; {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) &lt; 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}}  \cr    &amp; {p^2} &lt; \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] &lt; 2  \cr    &amp; {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}}  \cr    &amp; p &lt; q &lt; \sqrt 2  \cr} $$  However, my result doesn't necessarily belong to ${\Bbb Q}$. I wonder how did he obtain the value of $q$ in the book.  EDIT: Feel free to point out if there are more appropriate tags for this question.
56856
First of all, $n$ is even (find that out by trying to see what kind of remainders you can possibly get when dividing everything by 4). So $n=2m$, $m\in N$.  $ x^2 + 8 = 3^{2m} \Rightarrow x^2 - 1 = 3^{2m} - 9 \Rightarrow (x+1)(x-1) = (3^m + 3)(3^m - 3) \Rightarrow x$ is odd.  That's all I've tried so far. I'd like to get some help. Thanks.
47021
I'm trying to understand each part of this completed proof that my professor did, here is my interpretation in parentheses, please advise as necessary.  Proof: Assume that $2^{1/2}$ is rational, then $2^{1/2}$ = $p/q$ for some integer $p$ and $q$ with $q \neq 0$ and $\gcd(p,q) = 1$  $$(p/q)^2 = 2 = p^2 / q^2 $$ $$\text{ (What was the purpose of squaring √2 and p/q?)}$$  $$p^2 = 2q^2$$ $$ \text{ (Is this basically just canceling out the q^2 on left side, and multiplying the right side?)}$$  2 | p^2 => (2 divides p^2 since it is a factor)  2 | p => (thus 2 divides p)  p = 2k for some integer k => (definition of even)  (2k)^2 = 4k^2 = 2q^2 => (Why are we squaring 2k, and how does 4k^2 = 2q^2?)   q^2 = 2k^2 => (Canceled out 2 on one side, and dividing on other)  2 | q^2 => (2 divides q^2)  2 | q => (thus 2 divides q)  2 | gcd(p,q) => (2 divides gcd of p and q)  2 | 1 Which is a Contradiction (no clue how I got here)
60673
Firstly, I apologise that this question is specific to two polynomials. I understand that this post will not help a lot of people, and for that I am sorry. I will make it up to you all by anwsering lots more questions in the future! However, as my lecturers enjoy time off on the weekend, I am unable to ask this to anyone else who would know what the heck I'm talking about. Sorry if my notation is not widely used, I can elaborate more if required. Please excuse my poor Latex too.  Let:  $B = Y^2 - X^3 + X$, and   $F = (X^2 + Y^2)^3 - 4X^2Y^2$.  I wish to find the intersection number of $B$ and $F$ at $P = (0,0)$  i.e $I(P,F \cap B). $  However, I want to use the following proposition to do this:  If $P$ is simple on $B$ then $I(P,F \cap B) = ord_P^B(F) $.  I will omit proving that $P$ is simple on $B$ as I hope it is clear to see it is.  I know that $I(P,F \cap B) &gt; M_p(F) * M_p(B) = 1 * 4 $, where $M_P(G)$ is the multiplicity of $P$ on $G$.  Now, consider the Image of $X$ and $Y$:  $K[X,Y] \longrightarrow \frac{K[X,Y]}{Y^2 - X^3 + X} \longrightarrow O_P(Y^2 - X^3 + X)$  $X \longrightarrow x$  $Y \longrightarrow y$  Where lower case denotes the image.  Then $ord_P^{Y^2 - X^3 + X}((X^2 + Y^2)^3 - 4X^2Y^2)$  Can be found by substituting $y^2 = x^3 - x$ into $(x^2 + y^2)^3 - 4x^2y^2$  Which gives:  $ord_P^{Y^2 - X^3 + X}(x^9+3x^8 -5x^6 -4x^5 +3x^4 +3x^3)=3$  i.e $I(P,F \cap B) = 3$, a contradiction to  $I(P,F \cap B) &gt; 4$ as above.  The correct answer I know is 6, so where have I gone wrong?
73489
I know how to prove the fact that $ p$ | ${ p \choose k } $ (when writing it as a fraction, $p$ cannot be divided by any of the $1\times2\times...\times k$ or $1\times2\times...\times(p-k)$ because $p$ is prime).  When I try to apply the same rationale for $ p$ | ${ p^n \choose k } $ I get stumped because $p^n$ is in fact divisible by one of the $k! = 1\times2\times...\times k$, where $0 &lt; k &lt; p^n$.  If we take the example of $2^3$ and $k = 7$, we can easily see that $k! = 1 \times 2 \times ... \times 6 \times 7$, and $2$ clearly divides $2^3$.  How else can I approach this?
5350
I am pretty sure this question has something to do with the Least Common Multiple.    I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$. If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove. But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.
14947
Find all pairs of prime nummbers $p,q$ such that $p^3 - q^5 = (p+q)^2$.  It's obvious that $p&gt;q$ and $q=2$ doesn't work, then both $p,q$ are odd. Assuming $p = q + 2k$ we conclude, by the equation, that $k|q^3 - q - 4$ because $\gcd(k,q)=1$ (else $p$ is not prime) and $k=1$ has no solution.  I also tried to use some modules, but I couldn't.
113867
Why $3$ and $7$? I know from some reading that Hurwitz's Theorem explains this, but can someone help me build some intuition behind this or perhaps provide a simpler explanation? It still seems mysterious to me.
39690
I think this is true because of prime factorisations, i.e.  If $3$ a factor of the prime factorisation of $q^3$, then $3$ is a factor of the prime factorisation of $q$.  Therefore If $3$ divides $q^3$, then $3$ divides $q$.  Or for a general case, if $n$ divides $q^3$; then $n$ divides $q$. Is this correct?
100755
If  $x^2 \equiv y^2 \pmod {p^r}$ then $p^r \mid x^2 - y^2$ and so $p^r\mid(x-y)(x+y)$. Now since $p$ is a prime that is not dividing $x$ nor $y$ then it's easy to see that  $p^r \nmid x$ and also $p^r \nmid y$. Now I was wondering if this implied in particular that $p^r$ doesn't divide any linear combination of $x$ and $y$ ?  But This is certainly not true. Because we have that $3 \nmid 5$ and $3 \nmid 7$ however we have that $3 \mid 5 + 7 =12$.   I wanted to apply Euclids lemma but now I am stuck here.
21934
The most common way to prove the irrationality of a number is to do it by  contradiction. However, in this proof the contradiction arises from having two numbers, say p\q that are assumed to not have any common factors but are later proved to share the number under the radical as a factor. But for instance if the question so arises:  Is √4 rational or not? Answer with a proof.  At first glance, it is obviously rational, but assuming I do not know that √4 can be simplified to 2, what would stop me from arriving at irrationality as my conclusion?   This question seems quite trivial and probably stupid. Is this meant to be by definition or am I missing something fundamental to the aforementioned fallacy?
91740
Rationalize   $\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$  How would I go about doing this without wading through lots of algebra?  Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?  Thanks in advance!
65136
I am just looking for some clarification on this exercise:  Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.  I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:  $R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$  Here $R$ is reflexive, symmetric, and not transitive.  What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.  Thanks,  Tony
78054
Is the following correct way of showing that there is no simple group of order $pq$ where $p$ and $q$ are distinct primes?  If $|G|=n=pq$ then the only two Sylow subgroups are of order $p$ and $q$.  From Sylow's third theorem we know that $n_p | q$ which means that $n_p=1$ or $n_p=q$.  If $n_p=1$ then we are done (by a corollary of Sylow's theorem)  If $n_p=q$ then we have accounted for $q(p-1)=pq-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.  Is that correct?