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 ##.
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, ##...
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...
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?
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...
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."
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)\\...
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}\\...
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}...
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) ##...
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...
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.
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...
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...
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 ##?
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...
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\\...
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?
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...