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But how should I verify that ## N\equiv 1\pmod {6} ##? And I think I made some mistakes in my previous proof attempts, because ## p\mid N ## and ## p\mid (2p_{1}\dotsb p_{n})^{2} ## implies that ## p\mid (N-(2p_{1}\dotsb p_{n})^{2}) ##, so ## p\mid 3 ##.

I apologize for the confusion. What I meant is the symbol for expressing Legendre symbol, not divisibility.

Because ## 3\nmid 2p_{1}p_{2}\dotsb p_{n} ##. And ## p\mid N, p\mid 3 ## implies ## p\mid (N-3) ##, so ## p\mid (2p_{1}p_{2}\dotsb p_{n}) ##. Also, how should I show that ## N ## has an odd prime divisor of the form ## 6k+1 ##? At first, I thought this is so because ## N ## itself is odd. But it...

Okay, so I revised this proof: Suppose for the sake of contradiction that the only primes of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##. Consider the integer ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3=(2p_{1}p_{2}\dotsb p_{n})^{2}+3 ##. Since ## N ## is odd, it follows that ## N ##...

Proof: Suppose that the only prime numbers of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##, and let ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3 ##. Since ## N ## is odd, ## N ## is divisible by some prime ## p ##, so ## 4p_{1}^{2}\dotsb p_{n}^{2}\equiv -3\pmod {p} ##. That is, ##...

Thank you for the clarification, @andrewkirk @topsquark .

Does the above quantifier represent/symbolize that all of the integers ## a, b, c, d ## cannot be ## 0 ##? Is this correct?

So now we have that ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ## for all ## k\{8, 13, 20\} ## and ## \chi(k)^{6}=\chi(k^{6})=\chi(1)=1 ## for all ## k\{2, 4, 5, 10, 11, 16, 17, 19\} ##. And this implies that ## \chi(n)=8, 13, 20 ## can either be ## -1, 1 ##. But...

## 1, 8, 13, 20 ## ## 1^{2}\equiv 1\pmod {21}, 8^{2}\equiv 1\pmod {21}, 13^{2}\equiv 1\pmod {21}, 20^{2}\equiv 1\pmod {21} ##

Since ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. So ## G ## can be generated by order ## 2 ## or ## 6 ##. And we have ## \chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1 ##...

No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of ## G ## is isomorphic to ## C_{6}\times C_{2} ##, which are the two cyclic groups of order ## 6 ## and ## 2 ##. But how did you get these?

I already did. But I still don't understand. How should I find normal subgroups?

This is my question, too. I do not know what they mean, I just posted them under the relevant equation(s) just because my book has these definitions. Since these definitions are preventing people to make sense of my question, then please ignore them. How should I find those values then, starting...

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 1 & 2 & 4 & 5 & 8 & 10 & 11 & 13 & 16 & 17 & 19 & 20 \\ \hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \chi_{2}(n) & 1 & -1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\ \hline \chi_{3}(n) & 1 & 1 & 1 & -1 & 1...

I wish there are more details in this question, but no. The question states: "Write an account of prime numbers in arithmetic progressions. Your account should be in the form of an essay of 500-1000 words."

I am not completely sure of whether or not this will work for my purpose.

How should I write an account of prime numbers in arithmetic progressions? Assuming this account should be in the form of an essay of at least ## 500 ## words. Should I apply the formula ## a_{n}=3+4n ## for ## 0\leq n\leq 2 ##? Can anyone please provide any idea(s)?

After breaking down into smaller pieces, I got the following: \begin{align*} &(y'(x)+\tau\psi'(x))^{2}=y'^{2}(x)+2y'(x)\tau\psi'(x)+\tau^{2}\psi'^{2}(x)\\ &\omega^{2}(y(x)+\tau\psi(x))^{2}=\omega^{2}y^{2}(x)+2\tau\psi(x)\omega^{2}+\omega^{2}\tau^{2}\psi^{2}(x)\\...

Do you mean ## S(y+\tau\psi)=\int_{a}^{b}\frac{d}{d\tau}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx ##? If so, then how to evaluate this?

I am not sure if this is correct, but here is my work by using the definition of the Gateaux differential: \begin{align*} &dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\...

So both ## \pi(x)\approx \frac{x}{\log {x}} ## and ## li(x)\approx \frac{x}{\log {x}} ##? How did you get ## (1+O(\log^{-1} (x))) ##?

Thank you for pointing that out on part a). Also, another part of this question asks to prove that ## g(x)\sim\frac{1}{4}x ## by assuming that ## f(x)\sim\frac{1}{4}\pi(x) ##. By definitions, both ## \pi(x)=\sum_{\substack{prime p\leq x}}1 ## and ## v(x)=\sum_{\substack{prime p\leq x}}\log {p}...

How did you get ## g(40)\approx 6.8 ##? I thought it's ## g(40)\approx 2.95 ##.

a) ## f(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}1=3+13+23=39 ## ## g(40)=\sum_{\substack{prime p\leq x \\ p\equiv 3\pmod {10}}}\log {p}=\log {3}+\log {13}+\log {23}=\log {897} ## b) Proof: Let ## f(n)=\log {n} ## and ## a_{n}=1 ## if ## n\leq x ## is prime such that ##...

Just to confirm from where you left off, \begin{align*} &\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\int_{2}^{x}(\frac{1}{t\log {t}}\cdot \frac{1}{\varphi(k)}+\frac{R(t)}{t\log^2 {t}})dt\\ &=\frac{1}{\varphi(k)}+O(\frac{1}{\log {x}}+\frac{1}{\varphi(k)}\int_{2}^{x}\frac{dt}{t\log...

The following proof below is from the book: Let ## f(x)=\frac{1}{\log {x}} ## and ## a(n)=\frac{\log {n}}{n} ## if ## n\equiv h\pmod {k} ## is prime and ## 0 ## otherwise. By Dirichlet's Theorem, we have ## \sum_{n\leq x}a(n)=\frac{1}{\varphi(k)}\log {x}+R(x) ##, where ## R(x)=O(1) ##...

Before I try to prove the theorem, I want to know, how did you get those numbers from ## 3 ## to ## 1553 ## from the sequence?

From the textbook and notes. But if ##f(t)## can be any on ##[x, y]##, then what would you choose? How to choose this ##f(t)## function wisely?

Before I apply/use the Abel's summation formula, how should I find ## f(x) ##?

Don't say sorry, it's okay. I am the one who should apologize, because many stuffs above are things I never learned, I have no knowledge in group theory. So these are all new to me. For example, ##A\trianglelefteq \mathbb{Z}_{23}^*## is a subgroup, I've never seen ## \trianglelefteq ## before...

I still don't really understand. Can you please tell me why?

So to rule out the following situation in which you listed above, one example would be ## S\cap B=S ## if ## B=S ##?

Aren't the two statements of ## A=S ## and ## B=T ## ("automatically") the same as proving what's already proven from the first decomposition: ##S\cdot S\subseteq S\, , \,T\cdot T \subseteq S\, , \,T\cdot S\subseteq T.##? What's different in the second decomposition from the first...

Proof: Let ## p ## be an odd prime and ## G=\left \{ 1, 2, ..., p-1 \right \} ## be the set which can be expressed as the union of two nonempty subsets ## S ## and ## T ## such that ## S\neq T ##. Observe that ## p-1=22\implies p=23 ##. Let ## g\in G ##. Since ## g ## is either an element of ##...

The answer is ## 0.5 ## if you enter ## \sqrt{\dfrac{16}{64}} ## into any online calculator. So I wouldn't say ## \frac{1}{2} ## is wrong. And ## \frac{4}{8}=\frac{1}{2} ##. It's the same thing.

What a long proof. But what does ## aG^{2} \mapsto ({\frac{\cdot }{p}}) ## mean/indicate? I've never seen this before.

Let ## a ## be a primitive root of ## p ##. Then the integers ## a^{1}, a^{2}, ..., a^{p-1} ## form a reduced residue system modulo ## p ## such that ## \varphi(p)=p-1 ##, where ## r\in\left \{ 1, 2, ..., p-1 \right \} ##. This implies ## r\equiv a^{k}\pmod {p} ## for ## 1\leq k\leq p-1 ##. By...

I was thoughtless.

From ## \frac{p-1}{2} ##, to find the number of quadratic residues and non-quadratic residues.

That seems to be right. Because ## 73 ## has 36 quadratic residues and 36 non-quadratic residues. So they should cancel each other out leaving the answer to ## 0 ##. Also, earlier you mentioned that there's another theorem which claims ## \sum_{r=1}^{p-1}(r|p)=0 ## for any prime, where both the...

So ## \sum_{k=1}^{p-1}k\cdot (\frac{k}{p})=[1+72\cdot (\frac{72}{73})]+[2+71\cdot (\frac{71}{73})]+\dotsb +[36\cdot (\frac{36}{73})+37\cdot (\frac{37}{73})]=73+73+\dotsb +73 ##?

Maybe by substituting ## (r|p)=(-1)^{(p-1)/2}(p-r|p) ## into ## \sum_{r=1}^{p-1}r(r|p)=0 ##? But then we have ## \sum_{r=1}^{p-1}r(r|p)=\sum_{r=1}^{p-1}r(-1)^{(p-1)/2}(p-r|p) ##. How should I proceed from here and simplify to ## \sum_{r=1}^{p-1}r(r|p)=0 ##?

Since ## p=73 ## in this problem, how should I prove that ## \sum_{r=1}^{73-1}r(r|73)=0 ##? Given that ## 73=1\pmod {4} ##.

Proof: Suppose for the sake of contradiction that the only primes of the form ## 8k-1 ## are ## p_{1}, p_{2}, ..., p_{n} ## where ## N=16p_{1}^2p_{2}^2\dotsb p_{n}^2-2 ##. Then ## N=(4p_{1}p_{2}\dotsb p_{n})^2-2 ##. Note that there exists at least one odd prime divisor ## p ## of ## N ## such...

I found my mistakes.

Let ## x ## denote the number of sparrows, ## y ## denote the number of turtle doves and ## z ## denote the number of doves. Then we have ## \frac{1}{3}x+\frac{1}{2}y+2z=30 ## such that ## x+y+z=30 ##. Observe that \begin{align*} &\frac{1}{3}x+\frac{1}{2}y+2(30-x-y)=30\\...

Thank you for this!

And I have another question, since you said that ## 92 ## is therefore not an upper bound for possible values ## p ##, shouldn't the answer be infinitely many primes ## p ##? Don't we have infinitely many prime numbers?

That's what I was thinking earlier too. Sorry for the late response. I see it now.

From the second case, I found out that ## (7|23)=(11|23)=(19|23)=-1 ## because ## 23\equiv 3^{2}\pmod {7}, 23\equiv 1^{2}\pmod {11}, 23\equiv 2^{2}\pmod {19} ## from solving ## 23\equiv x^2\pmod {p} ## for all ## p\in\left \{ 7, 11, 19 \right \} ##. But I do not see where you lost the solutions...