# Tag Info

## Hot answers tagged number-theory

11

My interpretation of the given answer: We see that $222222$ is divisible by $13$. There's two tools in use here: $$222222[\text{some digits}] \equiv [\text{some digits}] \pmod {13}$$ and $$[\text{some digits}]222222$$ is divisible by $13$ if and only if $$[\text{some digits}]$$ is divisible by $13$. This means, in order to determine ...

7

An overkill proof. By the Lindemann--Weierstrass Theorem, $e^i$ is transcendental. I can get $e^i$ from $\sin(1) = (e^i-e^{-i})/(2i)$ by solving a quadratic equation. Thus, if $\sin(1)$ were algebraic, then also $e^i$ would be algebraic.

4

For $k\in\mathbf{N}^{*}$ I note $k = \prod_{p\in\mathscr{P}} p^{v_p(k)}$ the unique decomposition of $k$ in product of prime numbers. Then you have by the chinese reminder theorem, and as a product is a special kind of projective limit and as projective limits commute (i.e., can be interverted) : $$\varprojlim_{n\geq 1} \mathbf{Z} / n \mathbf{Z} \simeq ... 4 Note that k!+2,k!+3,\ldots,k!+k ends with k!+k which suggests that the sum stops once you reach k. So, for k=2, this means that the 2!+3 is not included. I do think that the question could have been better written as, "Let k\geq2 with k\in\mathbb{N}. Show that each k!+i, with 2\leq i\leq k, is composite." In order to answer this question, ... 4 I would recommend a series of books, specifications, libraries and CAS programs. Books A Course in Number Theory and Cryptography, Neal Koblitz (very dense, but an amazing book) An Introduction to Mathematical Cryptography, Jeffrey Hoffstein, Jill Pipher, J.H. Silverman (very readable and excellent book, which is more up-to-date) An Introduction to ... 3 The finite extensions of \Bbb Q_p are also local fields, so you can use the usual characterization of local field extensions being unramified. Recall: a finite extension of local fields L/K is unramified iff [L:K]=[\lambda:\kappa] where \lambda,\kappa are the residue fields for L,K respectively. This comes from the fundamental relationship that ... 3 The norm is the product of the conjugates, in the sense of algebraic extensions. The conjugate of \omega is the other root of its minimal polynomial x^2 + x + 1 – it's \omega^2. So$$N(a + b \omega) = a^2 - ab + b^2.$$Now to have N(a + b \omega) = 1, either a = 0, so we obtain \pm \omega as units. Or a \ne 0, and we can write a^2 - ab + b^2 ... 3 One can get some information with basic geometry, using the fact that this equation corresponds to a conic plane curve. There are only two possible answers, and they are exact (no error terms); deciding which answer is correct seems harder. Let me write my thoughts in case they are helpful to you. Your F is a homogeneous polynomial of degree 2, ... 3 The prime number theorem says that the n-th prime is p_n \sim n \log n. So your sum becomes$$ \sum_{p\le x}\frac{1}{p^{\alpha}} \sim \sum_{n : p_n \le x}\frac{1}{n^\alpha (\log n)^\alpha}\sim\sum_{n=2}^{x/ \log x}\frac{1}{n^\alpha (\log n)^\alpha} \sim \int_{2}^{x/\log x}\frac{dn}{n^{\alpha}(\log n)^\alpha} \sim \frac{n^{1-\alpha}}{(1-\alpha)(\log ...

3

You have a mistake in your first equality: $(x+y)^{p^{n+1}} = (x+y)^{p^n \cdot p}$, while $(x+y)^{p^n}(x+y)^p = (x+y)^{p^n+p} \neq (x+y)^{p^{n+1}}$. Correcting this mistake fixes everything: \begin{align*} (x+y)^{p^{n+1}} &= (x+y)^{p^n \cdot p} = ((x+y)^{p^n})^p = (x^{p^n} + y^{p^n})^p = (x^{p^n})^p + (y^{p^n})^p = x^{p^{n+1}} + y^{p^{n+1}} \, . ...

3

Every nonzero element of the ring of $p$-adic integers $\mathbb{Z}_p$ can be written in a unique way as $p^mu$, where $u$ is invertible in $\mathbb{Z}_p$. The field $\mathbb{Q}_p$ is the field of fractions of $\mathbb{Z}_p$, so its nonzero elements are of the form $$\frac{r}{s}$$ with $r,s\in \mathbb{Z}_p$, $r\ne0$. Then $$r=p^mu,\quad s=p^nv$$ so $$... 3 Hint: Consider 4(x^2 +x+1) = (2x+1)^2 + 3. Therefore, if it has a solution modulo p, -3 must be a quadratic residue. Use quadratic reciprocity. 3 First, recall that x^m-p is always irreducible over \mathbb{Q} for p prime (via Eisenstein's criterion). Consider the field \mathbb{Q}[\sqrt[n]{7},\sqrt[n+3]{7}]. It contains \mathbb{Q}[\sqrt[n]{7}] and \mathbb{Q}[\sqrt[n+3]{7}] so its dimension is divisible by n and n+3. They are coprime (since 3\nmid n) so the dimension is at least ... 3 If there are r\ge1 odd primes that divide n, then 2^r divides \phi(n) (*). And if 2^t (t\ge1) divides n then 2^{t-1} divides \phi(n). Therefore, if \phi(n)=14, then n is the power of an odd prime (n=p^s) or its double (n=2p^s). Any case, \phi(n)=p^{s-1}(p-1) so we have two possibilities: If s>1 then p^{s-1} divides ... 2 The only numbers with exact 2 proper divisors are the numbers of the form p^2, where p is a prime. The proper divisors are 1 and p in this case, and p+1 with p prime can only be a perfect square for p=3. This follows from the equation p=a^2-1=(a-1)(a+1). If a>2 , then p cannot be a prime. So, there is only 1 case of 2 ... 2 You will find a lot of intuitive reasons why Fermat last theorem holds in the following book : "Modular Forms and Fermat’s Last Theorem", from Cornell, Silverman and Joseph, I quite liked it in my young years. 2 Note: Sequences start with n=0. I. k = 2 The Lucas numbers,$$L_n = x_1^n+x_2^n = 2,1,3,4,7,11,18,29,\dots$$and the Fibonacci numbers,$$F_n = \frac{x_1^n}{y_1}+\frac{x_2^n}{y_2} = 0,1, 1, 2, 3, 5, 8, 13, 21,\dots$$where y_i =2x_i-1 and the x_i are the roots of x^2-x-1=0. Then,$$x_1^n = ...

2

A number ${d_{1}}\dots{d_{80}}$ is divisible by $13$ if and only if: ${d_{1}}{d_{2}}-{d_{3}}{d_{4}}{d_{5}}+{d_{6}}{d_{7}}{d_{8}}-\dots+{d_{78}}{d_{79}}{d_{80}}$ is divisible by $13$. We know that $-{d_{45}}{d_{46}}{d_{47}}+{d_{48}}{d_{49}}{d_{50}}-\dots+{d_{78}}{d_{79}}{d_{80}}=0$. We also know that ...

2

For the first time, it is an application of Eisenstein's criterion. For the second part, it is more or less correct, but you should consider that the Legendre's symbol is defined for integers, and $\mathbb{Z}_p$ is a subset of $\mathbb{Q}_p$: if $(x/y)^2 \equiv a \pmod{p}$ then $p$ cannot divide $y$, so there exists a multiplicative inverse $z\neq 0$ such ...

2

The question boils down to the following: How many $x_i \in \{ 5,7 \}$ can you find such that $$x_1...x_{19}5 \equiv 0 \pmod{7}$$ Since $7$ is divisible by $7$, we can replace all $7's$ by $0$s. Define $y_i =1$ if $x_i=5$ and $0$ otherwise. So the problem boils to $$5\cdot y_1...y_{19}1 \equiv 0 \pmod{7}$$ or $$y_1...y_{19}1 \equiv 0 \pmod{7} \,.$$ ...

2

Notice we are talking about $k!+n$ with $n\leq k$ which was not the case in the example you gave. Notice that in $k!+n$ with $n\leq k$ we can factor $n$ and therefore is composite

2

The following generating function enumerates every 20-digit number using only the digits $5$ and $7$ and ending in a $5$. $$G(x)=(x^5)(x^{50}+x^{70})(x^{500}+x^{700})\cdots(x^{5\cdot10^{19}}+x^{7\cdot10^{19}})$$ That is, $G(x)$ is the sum of $x^N$ for all such integers $N$. If we reduce each exponent in this polynomial modulo $7$ and call the resulting ...

2

Turning a comment into an answer (of sorts).... Various papers (e.g., this one by Yann Bugeaud and Preda Mihailescu) cite papers by Nagell and Ljunggren as proving there are no perfect squares of the given form other than the ones the OP found. As for the more general problem of representing an arbitrary perfect power, a recent paper by Michael Bennett and ...

2

Assume $\gcd(a,n)=1$. Define $c$ by $ac\equiv1\bmod n$. Then $$aj^2+2jk+bk^2=a(j^2+2cjk+bck^2)=a((j+ck)^2+(bc-c^2)k^2)$$ so the sum is $$\sum_ke(a(bc-c^2)k^2)\sum_je(a(j+ck)^2)$$ This solves your problem about things not being integers.

2

(Old answer revised with new information.) Given, $$d(2^{k+1}-1)-b^2(2^{k+1}-2)=1$$ There are solutions for all $k$. It is best to simplify with $k=p-1$ to get $\color{brown}{M_p = 2^p-1}$. Hence, $$d(M_p)-b^2(M_p-1)=1\tag1$$ I. Family 1: The two general solns in the old answer can be combined into one. For any $p$, there is an infinite number of ...

2

Recall again the prime number theorem, which gives that $$p_n \approx n\log n.$$ Then the sum we want can be compared to $$\sum_p \frac{1}{p \log \log p} \sim \sum_n \frac{1}{n\log n \log \log n}.$$ I claim this latter sum diverges based on the integral test for convergence. Claim: The following series diverges: $$\sum_n \frac{1}{n \log n \log \log ... 2 Long story short,$$ p_n \leq C n \log n $$by Chebyshev's theorem (\pi(n)\geq D \frac{n}{\log n}), hence the series is lower bounded by a multiple of$$ \sum_{n\geq 3}\frac{1}{n\log n \log\log n}$$that diverges due to Cauchy condensation test (applied twice). 2 It suffices to prove that \ln^2x < x for all x>1, since this gives you$$\sum_{n=1}^N \frac1{\ln^2 p_n} > \sum_{n=1}^N \frac1{p_n} \stackrel{N\to\infty}\longrightarrow \infty$$To show this see that \ln^2 1 = 0 < 1 and$$\frac{\mathrm d}{\mathrm dx} \ln^2 x = 2\frac{\ln x}x < 1$$2 With ordinary integers: Theorem of Gauss that x^2 + y^2 + z^2 integrally represents all positive integers except those$$ 4^k (8 n+7).  Meanwhile, if $x^2 + y^2 + z^2 \equiv 0 \pmod 4,$ then $x,y,z$ are all even. Put these together, we say that $x^2 + y^2 + z^2$ is anisotropic in $\mathbb Q_2.$ It is also anisotropic in the real numbers, as the sum ...

2

Yes this is proved by S.W.Golomb. I cannot resist to show a piece of my work:http://arxiv.org/abs/1311.1398

Only top voted, non community-wiki answers of a minimum length are eligible