# Tag Info

### Approach to Fermat's Last Theorem using group theory

Your strategy cannot work. For any prime $p$ and large enough $m$, $(**)$ will fail for $G_m$, that is, there will be $g\in G_m$ such that $g+1$ in $G_m$. Through your argument in point 4, this ...
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### Is the product of primes from $n/2$ to $n$ always less than $2^n$?

Let $P_n:=\prod_{p\leq n}p$ denote the product of all primes no greater than $n$. Then $$\log P_n=\sum_{p\leq n}\log p=:\theta(n),$$ where $\theta$ denotes the Chebyshev function. Then your product ...
• 65.1k
Accepted

### Let $p_n$ denotes the $n$-th prime. There are at least $p_n- 1$ primes between $p_n$ and $\prod_{k=1}^n p_k$·

Maybe a simpler solution would be to use so called "Bertrand Theorem" that is that there is at least one prime between $n$ and $2n$. That is, check the inequality for $n=2$" and then ...
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1 vote

### Let $p_n$ denotes the $n$-th prime. There are at least $p_n- 1$ primes between $p_n$ and $\prod_{k=1}^n p_k$·

Note that for $n=1,2$ the statement fails, so we only consider $n\ge3$. We can check manually that it holds for $n=3$ and assume from now on that $n\ge4$. By utilizing the bound (see a short and ...

### broader meaning of twin prime constant?

If $a_2 - a_1 = 2$ you are essentially counting some prime twins in a given interval. Consider $i > \sqrt n$ If $i = a_1 \mod p$ for all $p< \sqrt n$ with $a_1 \neq 0$ then $i$ is a prime. Then ...
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### Understanding a bound on number of primes congruent with $a$ mod $q$

By the definition of $\theta,$ as $t$ increases the expression for $\theta(t,q,a)$ jumps up by the amount $\log t$ each time when $t$ crosses another prime that satisfies the congruence condition. ...
• 2,460
1 vote
Accepted

### Reasoning about reduced residue systems as a generalization from prime gaps

Let $n<x$ be any natural numbers. Then the ordered list of natural numbers relatively prime to $p_n$ is $(1,2,\dots)$ with the multiples of $p_n$ removed. Thus the $c_{n,i}$'s divisible by $p_x$ ...
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### When and how does a rational prime (not ideal!) reduce in a given number field?

The answer to your question is indeed not easy and is subject to still ongoing research, compare for example http://faculty.fairfield.edu/pbaginski/Papers/FactoringAlgIntegersMonthly.pdf The reason is,...
1 vote

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### Are there any primes that satisfy the third iteration of this process?

The smallest sprime$_3$ is \begin{align*} 31571389&=15785689+15785699+1\\ &=(7892831+7892849+1)+(7892849+7892873+1)+1 \end{align*} where \begin{align*} 7892831&=3946403+3946421+1\\ 7892849&...
• 85.1k
1 vote
Accepted

### $\Delta U = U - U$ elementwise = $2R$ for every UFD $R$ and its group of units $U$. Or there is some formula...

Does this mean as well that $\Delta U \simeq \prod_{i\in I} \Delta U_i'$ as sets? Yes. It helps to think about elements of $U \simeq \prod_{i\in I} U_i'$ as vectors of length $|I|$ with ordinary ...
• 10.9k
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### Gamma is multiplicative with Zeta - Is there an additive version?

Yes $$I(s)=2\sum_{n\in \Bbb N} \int_{J=(0,n^{-s/2})} e^{\frac{\log^2({n^{s/2}})}{\log t}}~dt$$ and $$f(s)=\sum_{n\in \Bbb N} \log n~K_1(s\log n)$$ where $K_1$ is the modified Bessel function. Here's ...
• 104

### The square roots of different primes are linearly independent over the field of rationals

I would like to record another solution, close to the one given by Bill Dubuque. The goal is to show that $$\hbox{dim}_{\bf Q} {\bf Q}[\sqrt{r_1},..., \sqrt{r_n}] = 2^n$$ if no $r_i$ or products of ...
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Experimental data suggests that the product converges, but to a value less than $4/\pi$. I computed the product for the first $1.8 \times 10^9$ primes. The results are as follows: Upto 2 \times 10^... • 21.5k 2 votes ### Understanding partial summation with primes numbers Here is my attempt and I obtain the same result as you (modulo the sign): by putting $$S(y) := \sum_{p<y} \frac{a_p}{p} = -\log(y) + \mathcal{O}(1),$$ we have \begin{align*} \sum_{p<\sqrt{x}} \... • 341 1 vote Accepted ### Carmichael Numbers sample test data Note that(6k+1)(12k+1)(18k+1)$is never prime. You want to check if each of the factors is prime. Here’s some PARI/GP code to generate a Carmichael number of the desired form larger than a given ... • 32.4k 1 vote ### How random is the modulus of a large semiprime number? You’ve already observed that the semi prime will be odd, and you can use this to shave off one bit. You can do the same trick with the first few primes (I think the first 233 primes) where you store ... • 32.4k 3 votes ### Product formula for$\dfrac{4}{\pi}$As something of a ground rule, we might require the factors to be listed in order of increasing value of the prime, so that the different residues$\bmod 5$are mixed together and thus cancel out, by &... • 42.4k 0 votes ### Bounds for the Prime Zeta function By partial summation we have $$\sum_{p\leq x}\frac{1}{p^{s}}=\sum_{n\leq x}\frac{1_{p}}{n^{s}}=\frac{\pi\left(x\right)}{x^{s}}+s\int_{1}^{x}\pi\left(t\right)t^{-s-1}dt$$ so $$\sum_{p\in\mathbb{P}}\... • 32.8k 1 vote ### Finding Twin primes Such That Their Sum + 1 is Also Prime I can't provide a necessary condition for OP's query, but I can provide a sufficient one. If (p,p+2) are twin primes, then 2p+3 will be prime if (2p+1,2p+3) are also twin primes. This might seem ... • 7,523 3 votes ### What is the expected size of primes sampled uniformly at random from the set of first 2^n primes? Let x be the 2^nth prime. Then the expectation in question is precisely$$ \frac1{2^n} \sum_{p\le x} \log_2 p = \frac1{2^n\log 2} \sum_{p\le x} \log p.$$By the prime number theorem, the latter ... • 85.1k 1 vote ### How many coprimes to$n$are also prime? From the above comments it seems like OP wants a closed form answer. There are no known exact and useful closed formula for$\pi(x)$and$\omega(x)$. So you will not have a formula that gives you the ... • 21.5k 2 votes ### How many coprimes to$n$are also prime? If$\pi (x)$be the prime counting function and$\omega (x)$be the prime omega function (that counts the number of distinct prime factors) then what you want is$\pi (n) - \omega (n)$- this is ... • 97 0 votes ### If$a, b, c, d$are natural numbers, such that,$ab = cd$, prove that$a^2 + b^2 + c^2 + d^2$is a composite number. Consider,$a^2+b^2+c^2+d^2$by using completing the square method we get$(a+b)^2+(c-d)^2 =(a-b)^2+(c+d)^2$Where$ab=cd$,$a>b$&$c>d\$ Fermat's two square theorem : Every prime of the ...

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