# Tag Info

## New answers tagged number-theory

0

The sigma function also shows up naturally in the study of modular forms and elliptic curves, which makes it significant in algebraic geometry (hence physics) and in cryptography.

1

There is 1 way to select to do the same thing out of a set of activites (of which one could be "nothing") over and over $n$ times in a row if that is what you mean by "doing nothing". You could also consider this combination: $$\left(\begin{array}{c}n\\0\end{array}\right) = \frac{n!}{n!\cdot 0!} = 1 \text{ iff } 0! = 1$$ One combination to pick 0 out of ...

2

That's right. One way I feel is a good way to argue this is to look at two collections in conjunction with each other, say Box A and Box B. There are $n$ objects in Box A, and $m$ objects in Box B, and let's suppose we want to find all possible ways of arranging the items in Box A, and then in Box B, separately . Of course, $$N = n!m!$$ Now suppose there ...

0

Yes to both alternatives. That there is only one way to do nothing is a strong motivation to ensure that the factorial is defined in such a way that you have $0! = 1$. However, every question of "why is this true" in mathematics ultimately boils down to "it's a consequence of how things are defined". (presuming, of course, that we know the fact with full ...

0

From the combinatoral beginning it is like that by definition, but later developed branches of maths shows $Γ(n + 1) = n! \ \ n = 0, 1, 2...$

1

Just do them by cases: Case 1 $x$ natural; $y$ natural. $x + y$ and $xy$ are natural and therefore integers. If $x >y$ then $x -y$ is natural and there fore an integer. If $x = y$ then $x - y = 0$ and therefore an integer. If $x < y$ then $y-x$ is natural and $x - y = -(y-x)$, a negative of a natural and therefor an integer. Case 2 $y$ is 0. (the ...

4

Yes. 0! = 1 because it is defined that way. One of the reasons it is defined that way is because it makes sense in the context of combinatorics given that there is only one empty object or permutation up to isomorphism. So both are correct - it varies on one's definition of the factorial function. When defined in reference to a combinatorial quantity it ...

10

Yes, precisely there is a unique function $\emptyset \to \emptyset$ (with empty graph), which happens to be a bijection ($\operatorname{id}_\emptyset$). Note, that $n!$ is the number of bijections $\{1,\dots, n\}\to \{1,\dots,n\}$.

7

For positive numbers the factorial function $n!$ is defined as the product of all positive integers less or equal to $n$. To define $0!$ we need to "extend" the definition. Another way to define it is to notice that: $$(n-1)! = \frac{n!}{n}$$ Pluging $n=1$ we get: $0! = \frac{1!}{1} = 1$

0

The number of significant digits is the max number used in the equation. The best you could do is multiply a number by $1.33$ and get a number with 3 significant digits. The reason for this is the number of significant digits is a measure of accuracy of our estimated $1.33$ value. We don't really know if its closer to $1.327$ or $1.328$ in our calculation ...

2

You don't. Let $g$ be the "Goodstein terminating" function: $g(n)=k$ iff $G_k(n)=0$ and $G_{k-1}(n)\not=0$. Then $g$ grows insanely fast: if you've heard of the Ackermann function, it's of a similar species. In fact, new notation and concepts had to be invented to even talk about how fast such functions grow! Look up "fast-growing hierarchies" (or see ...

0

There are many algorithms on integer factorization, but they do not run in polynomial time (only sub-exponential time). So it is possible to determine the number of prime divisors of integers $n$ for a "reasonable size" on $n$. On the other hand, there is a lot known about solving the equation $\phi(n)=k$, where we do not need to know the number of prime ...

0

No, there is no known such computable function as the numbers get sufficiently large. For example, it's unknown how many prime factors googolplex+1 has, but it's over 14.

0

It tells you that $$a^{50}x_0 + c\left(\frac{a^{50}-1}{a-1}\right) \equiv x_{50} \mod m$$ And that is all. You'll need at least two more values of $x_n$ to determine what $a, c, m$ are.

1

Note that $a^p \equiv a \pmod p$ if $p \not \mid a$ and $a^p \equiv 0 \pmod p$ if $p \mid a$. This comes from Fermat's Little Theorem. So using these we can determine when a number is a p-remainder-number. So a number is p-remainder-number iff the sum of all digits of the number that aren't divisible by $p$ is divisible by $p$. For the second one I don't ...

1

In your setting, we have 1. $q$ is prime 2. $4q+1$ is prime 3. $6q+1$ is prime Your questions is whether $4q^2 +1$ is prime. Running the following program in SAGE, we find that $4q^2+1$ is not prime for $q=277$. for p in primes(100000): if is_prime(4*p+1)==1: if is_prime(6*p+1)==1: if is_prime(4*p^2+1)==0: ...

1

Let $f(n):=n+\sqrt{n}+\frac{1}{2}$ and $g(n):=\big\lfloor f(n)\big\rfloor$ for each $n=1,2,\ldots$. Observe that $f(n)$ is not an integer for any positive integer $n$. We have $1<f(n+1)-f(n)<2$, so that $$1\leq g(n+1)-g(n)\leq 2$$ for all $n=1,2,\ldots$. Suppose that $g(n+1)-g(n)=2$. Write $k:=g(n)$. Therefore, $k<f(n)<k+1$ and $k+2<f(n+1)... 1 Let$N = q^k n^2$be an odd perfect number with Euler prime$q$. In a recent preprint, Brown claims a complete proof for$q < n$, and a partial proof that the inequality$q^k < n$holds under many cases. (See arXiv.) In particular, since$q < q^k < n$holds if Brown's proofs are correct (and completed), then the resulting lower bound is $$\gcd(... 5 From the Prime number theorem, the prime numbers are asymptotically distributed according to$$ \pi(n) \approx \frac n{\ln n}\ , $$where \pi(n) is the number of primes less than or equal to n. This shows that for n=2^{2^{100}} we'd have$$ \pi(n) \approx \frac {2^{2^{100}}}{\ln 2^{2^{100}}}=\frac {2^{2^{100}}}{2^{100}\ln 2} \approx 2^{(2^{100}-99)}$...

1

$\frac{x}{ln(x)}$=number of primes until x (for large enough x, the error becomes almost nill) i think the rest is easy, btw search prime number theorem :)

1

The answer to your question is no. You can try to do the following exercise (taken from Serre, Local Fields, Ch. III, section 6): Define $A,B,K,L$ in the way you did above, and assume $B$ is 'completely decomposed', i.e., there are $n = [L:K]$ primes of $B$ above the prime $\mathfrak m$ of $A$. Then $B$ is of the form $A[x]$ (for some $x\in B$) if and only ...

0

$$44^2 - 1978 = -42 < 0$$ $$89^2 - 1978 \cdot 2^2 = 9 > 0$$

0

when does $\lfloor \sqrt n + \frac 12\rfloor$ jump? $(k+0.5)^2 = k^2 + k + 0.25$ when $n$ can be written as $k^2 + k$ you are on the low side of the next step, and when $n$ can be written as $k^2 + k + 1$ you are on the high side of a step. $\lfloor k^2 + k +\sqrt {k^2 + k} + \frac 12\rfloor = k^2 + 2k\\ \lfloor k^2 + k + 1+\sqrt {k^2 + k + 1} + \frac 12\... 3 There are$9\cdot10^4$five-digit numbers, since there are$9$choices for the first digit, and$10$choices for the next$4$digits. Of these numbers,$8\cdot 9^4$do not contain$4$as a digit. Therefore the number of$5$-digit numbers with at least one$4$as a digit is equal to $$9\cdot 10^4-8\cdot 9^4=37512$$ 1 The easiest is to just subtract all of the ones that don't: 1 digit numbers: 8 possibilities (1,2,3,5,6,7,8,9) 2 digit numbers: 8 possibilities for the first * 9 possibilities for the second (0,1,2,3,5,6,7,8,9) = 72 3 digit numbers: 8 possibilities for the first * 9 possibilities for the second (0,1,2,3,5,6,7,8,9) * 9 possibilities for the third (0,1,2,3,... 0 Since all the integers are positive, we can only have $$x_1+x_2+x_3=3$$ $$y_1+y_2+y_3+y_4=5$$ In the first equation, clearly$x_1=x_2=x_3=1$In the second,only one is equal to$2$and everything else is equal to 1 Hence, the solutions: $$(1,1,1)(2,1,1,1)$$ $$(1,1,1)(1,2,1,1)$$ $$(1,1,1)(1,1,2,1)$$ $$(1,1,1)(1,1,1,2)$$ 4 possibilities. 0$15 = 5 * 3$, So$(x_1 + x_2 + x_3) = 3$or$5$same for the other one.$(y_1 + y_2 + y_3 + y_4) = 3$has no solution.$\therefore(x_1 + x_2 + x_3) = 3$and$(y_1 + y_2 + y_3 + y_4) = 5$. Now you know$x_1, x_2, x_3$can't be greater than$1$.$\therefore$the only solution for it is$(1 + 1 + 1)$. In$(y_1 + y_2 + y_3 + y_4) = 5$, only one ... 3 Since$15 = 1\cdot15 = 3 \cdot 5$, the only possibilities are $$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (1, 15), (15, 1), (3, 5), (5, 3)$$ Since the variables are all at least$1$, we can rule out the case where$x_1 + x_2 + x_3 = 1$or where$y_1 + y_2 + y_3 + y_4 = 1, 3$for the minimum values of each sum exceeds the respective assignments. This ... -1 Suppose not. (We take the negation of the theorem and suppose it to be true.) Suppose there are three cube roots who are in arithmetic progression although not necessarily consecutively. Let our radicands be the primes$p, q,$and$r$and the difference between terms in the arithmetic progression be$d$.$k_1$and$k_2$respectively represent the number of ... 0 Despite your already mentioned error, I do not believe your question poser intended you to go through this route. If I understand correctly, your question poser did not necessarily want you to find all points.$f(x)=x^2is not one to one, in fact it is an even function so: $$f(x)=f(-x)$$ So it would be much easier if you try to find a rational distance ... 0 \begin{align}24u\equiv -2\pmod{17}&\iff 24u+2\equiv 0\pmod{17}\\&\iff 17\mid (24u+2)\\&\iff 17\mid 2(12u+1)\\&\iff 17\mid(12u+1)&\because\gcd(2,17)=1\\&\iff 12u+1\equiv0\pmod{17}\\&\iff 12u\equiv -1\pmod{17}\end{align} 2 The beauty of this problem is that it can be studied with any level of mathematics. Solving it however, is a different matter. The fact that so many have tried and failed proves the following: If proof with elementary mathematics is possible, then it requires some exceptional feat of logic, reasoning, or intuition, which has eluded so many people before. ... 0 I hope my answer is okay.6x+15y \equiv 9(mod 18)=> 2x+5y \equiv 3(mod 6)=> 2x-y \equiv 3(mod 6)$i.e$2x-y-3 \equiv 0(mod 2)$and$\equiv 0(mod 3)$this means$y+1 \equiv 0(mod 2)$and$x+y \equiv 0(mod 3)$Write$y=2p+1$and so$x-p +1 \equiv 0(mod 3)$Then use parametric equations. 2 Dividing everything by$3$, including the modulus, we get the equivalent congruence $$2x+5y\equiv 3\pmod{6}.$$ It is convenient to rewrite this as$2x-y\equiv 3\pmod{6}$, or equivalently$y\equiv 2x-3\equiv 2x+3\pmod{6}$. Now we have a parametric solution:$x\equiv t\pmod{6}$,$y\equiv 2t+3\pmod{6}$. To write it out at length, set$t=0,1,2,3,4,5$. The ... -2 Wolfram answer:$y = 2c + 1x = c + 3d + 2$2 If you want to eliminate multiples of 2 and 3, just work in base 6=2x3. The only possible primes are those that end in 1 or 5. Similarly, to sieve out numbers with all factors 2, 3, 5, and 7, work in base 2x3x5x7 = 210. This is standard stuff for those developing sieves, and there are many advanced takes on this. Read, play around, have fun, and don't ... 1 From $$(\sqrt{1978}+\lfloor \sqrt{1978}\rfloor)(\sqrt{1978}-\lfloor \sqrt{1978}\rfloor)=1978-\lfloor \sqrt{1978}\rfloor^2$$ one has $$y=\sqrt{1978}-\lfloor \sqrt{1978}\rfloor=\frac{1978-\lfloor \sqrt{1978}\rfloor^2}{\sqrt{1978}+\lfloor \sqrt{1978}\rfloor}\lt \frac{1978-\lfloor \sqrt{1936}\rfloor^2}{\sqrt{1978}+\lfloor \sqrt{1936}\rfloor}\lt\frac{1978-1936}{... 2$$\mathrm{ y=\{\sqrt{1978}\}=\{\sqrt{44^2+42}\} }$$Now suppose$$\mathrm{ 44^2+42=(44+z)^2\\ \implies y=\{44+z\}=\{z\}\cdots(I) }$$Also we have$$\mathrm{44^2+42<(44+1)^2\\ \implies 0<z<1}$$Therefore according to \mathrm{(I)} we have$$\mathrm{ y=z\\ \implies 44^2+42=(44+y)^2\\ \implies 2\cdot44\cdot y=42-y^2<42\\ \implies y<{42\over2\... 0 As you said,$k\cdot k^{-1} \equiv 1 \pmod p$. Therefore, whenever we have$k\cdot k^{-1}$in a$\pmod p$equation, we can replace it with$1$, since they are congruent in such a system. Here's an example: $$2x \equiv 3 \pmod{5}$$ Now, after doing some guess and check, you can find that$2\cdot 3 \equiv 1 \pmod 5$, meaning$2^{-1} \equiv 3 \pmod 5$. Notice ... 1 $$4\cdot 3=3+4+5$$ $$4\cdot 5=2+3+4+5+6$$ $$4\cdot 7=1+2+3+4+5+6+7$$ We are now at a roadblock. But we can continue by changing the method. $$4\cdot 9=1+2+3+4+5+6+7+8$$ $$4\cdot 11=2+3+4+5+6+7+8+9$$ $$4\cdot 13=3+4+5+6+7+8+9+10$$ $$4\cdot 15=4+5+6+7+8+9+10+11$$ etc., etc. Do you see how this method could be used to analyze other powers of 2? 2$8!+1$is divisible by$61$and$661$, neither of which is congruent to$1$mod$8$. See OEIS sequence A064164 and the paper of G. E. Hardy and M. V. Subbarao mentioned there. 0 As P Vanchinathan noted in a comment, you confused the coordinates of the points. Starting again from $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = {p\over q}$$ and now using the correct relationship$y_i=x_i^2$, we get $$\sqrt{(x_2-x_1)^2+(x_2^2-x_1^2)^2}=\frac pq\;,\\ (x_2-x_1)^2\left(1+(x_2+x_1)^2\right)=\frac{p^2}{q^2}\;.$$ If$x_1$and$x_2$are ... 1 Yes, they should say odd divisor$p \ge 3$. If$3|x$the progression can be$\frac x3-1, \frac x3, \frac x3+1$. Specifically, if$x=3$the term$\frac x3-1$is zero, which is not permitted, but we can just delete it and have the progression$1,2$This shows how to deal with any odd divisor$p$(which need not be prime). The progression is$\frac xp-\frac ...

3

A handful for $n$ odd 13 a: 230153 b: 12792 c: 283945 d: 284233 u: 507 v: 164 15 a: 32625 b: 3472 c: 61455 d: 61553 u: 56 v: 31 23 a: 523367 b: 57072 c: 1413145 d: 1414297 u: 984 v: 667 27 a: 206703 b: 3848 c: 231345 d: 231377 u: 156 v: 37 89 a: 11534489 b: 700920 c: 63439289 d: 63443161 u: 649 v: 540 105 a:...

3

Introduce for the RHS $$L_1(s) = \sum_{n\ge 1} \frac{2^{\nu(n)} \lambda(n)}{n^s}$$ and $$L_2(s) = \sum_{n\ge 1} \frac{\sigma_0(n)}{n^s} = \zeta(s)^2.$$ We have $$L_1(s) = \prod_p \left(1-\frac{2}{p^s}+\frac{2}{p^{2s}}-\frac{2}{p^{3s}} + \cdots \right) \\ = \prod_p \left(-1 + 2\frac{1}{1+1/p^s}\right) = \prod_p \frac{-1-1/p^s+2}{1+1/p^s} = \prod_p \frac{1-... 2 An Attempt You have a^2+\left(n^2+1\right)b^2=d^2. All integral solutions (a,b,d) to this equation satisfies (1) |a|=|d| and b=0, or (2) d\neq 0 and \left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right) for some r\in\mathbb{Q}. Now, if b^2+c^2=d^2, then (... 1 gcd(a,m)=1, then ax≡b \pmod m has exactly one solution x < m. There are infinitely many solutions, and each one is congruent to x \pmod m. 0 This can be done by means of Ramanujan Sum. We follow the definitions and notations in the link. The identity is a special case (s=2) of the following general identity:$$ \frac{\sigma_{s-1}(n)}{n^{s-1}\zeta(s)} = \sum_{q=1}^{\infty} \frac{c_q(n)}{q^s} \ \ \ (\ast) $$Proof of (\ast):$$ c_q(n)=\sum_{d|q} \mu\left(\frac qd\right) \eta_d(n) = \mu \...

1

That would be the same as Dicinson's Conjecture because assuming $a$, $b$, $n$, $n_2$, $n_3$... are pairwise coprime, there is some integer $x$ such that $ax+b$ divides $n_z$ by modular arithmetic. To show that there are infinitely 2, 3, 4 etc. almost-primes... would mean there are infinitely many primes of the form ($ax+b$)/$n$ (assuming $n_z$ is prime) and ...

1

No. The way the question is worded, $p=2$ also works. Anyway, $y \neq 0 \pmod p,$ this means $y$ has a multiplicative inverse $\pmod p,$ for no better reason than $\gcd(y,p)=1$ and we have integers with $ys+pt=1.$ $$x^2 + y^2 \equiv 0 \pmod p,$$ $$x^2 \equiv -y^2 \equiv 0 \pmod p,$$ $$\frac{x^2}{y^2} \equiv -1 \pmod p,$$  \left( \frac{x}{y} \right)^...

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