# Tag Info

1

You can just use the fact that: $$\sum_{i=1}^ni = \dfrac{n(n+1)}{2}$$ and take the nth root.

1

$$\sqrt[n]{1+2+3+\cdots+(n-1)+n}=\sqrt[n]{\frac{n(n+1)}2}$$

0

If you accept subrings of the complex numbers as numerical rings, then just take a homomorphism $\mathbb Z[X] \to \mathbb C$ with nontrivial kernel. This reduces to sending $X$ to an algebraic number. A prime example is $X \mapsto i$, which gives you the numerical ring $\mathbb Z[i]$. If you mean encoding the ring operations of $\mathbb Z[X]$ as integer ...

0

The requirements are a little vague, but it sounds like you'd like to find a function from a polynomial ring $R[X]$ into the set $\Bbb Z$, and then equip $\Bbb Z$ with a potentially unusual addition and multiplication to make the map a ring isomorphism, thereby "representing the ring $R[X]$ with integers." Further, the addition and multiplication are ...

0

To add pessimism in finding such prime $p$ that cannot be written as sum of $3$ co-divisors of cube, I've drawn images: Number of ways to write an integer number $p$ as the sum $x+y+z$, where $x\cdot y\cdot z=c^3$. (images might be slightly zoomed in) $$p\le 800$$ $$p\le 8\;000$$ $$p\le 800\;000$$ These images were made in style like here. (Since ...

1

Hint: $$(1+a+a^2)(a-1) \equiv a^3-1\equiv 0$$

0

We know, ord$_ma=d,$ ord$_m(a^k)=\dfrac d{(d,k)}$ (Proof @Page#95)

1

As you can also read in the comments to your question, the set of Gaussian rationals is a field. Seeing that a field is always a unique factorization domain, the answer to your question is yes!

2

Let $a=4$, $b=8$ in $4{\bf Z}$.

8

Let $\,a = 2,\ b=\sqrt{12}\,$ in $\,\Bbb Z[\sqrt{12}]$

3

In this answer, I give a pictorial proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$ With your sum after dividing by $2$ we have $$\frac14+\frac38+\frac5{16}+\frac{7}{32}+\frac{9}{64}+\cdots$$ and the same picture can be used if we expand the original $\frac12\times\frac12$ square to the left as well as to the right and upward. The ...

3

$$\sum_{n=1}^{+\infty}\frac{2n-1}{2^n}=-1+2\sum_{n=1}^{+\infty}\frac{n}{2^n}=-1+2\sum_{n=1}^{+\infty}\sum_{m\geq n}\frac{1}{2^m}=-1+4\sum_{n=1}^{+\infty}\frac{1}{2^n}=-1+4=3.$$

3

Another approach is to use summation by parts. This method works as follows. If $$S_N = \sum_{n=1}^{N} a_n b_n$$ then we can define $$B_n = \sum_{k=1}^{n} b_k$$ Then $$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$ In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then $$B_n = \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$$ and a_{n+1} ... 2 \begin{align} \sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots \\[1ex] 2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}} \\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j} \\ &= 1 + \sum_{j=1}^\infty \frac {2j+1}{2^j} ... 1 We can take this as the x=\frac12 case of the seriesf(x)=x+3x^2+5x^3+7x^4+\cdots+(2n-1)x^n+\cdots = \sum_{n=1}^\infty (2n-1)x^n$$We can recognize in this summation the geometric series$$S(x)=x+x^2+x^3+\cdots= x+x\left(x+x^2+\cdots\right)=x+x S(x)=\frac{x}{1-x}$$where in the last equality we have solved the rest of the equation for S(x). The ... 7$$\frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots=\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$$4 Your proof is basically valid, but there are special circumstances for this particular ring of (algebraic) integers. You first need to check that y can't be divisible by 3. Note that ( in the usual ring of integers), if we have 3 divides 1 + x + x^{2}, then we must have x \equiv 1 (mod 3),say x = 3z+1 for some integer z. Then x^{2}+x +1 = ... 0 I believe I recall reading the name "solvable closure" for this construction, but Google doesn't seem to turn up anything. This would make sense, however, as F(f) is the smallest extension of f which is closed under the operation of taking solvable extensions. Alternatively, \text{Gal}(F(f)/f) is the largest pro-solvable quotient of ... 0 As far as I can tell — with the qualification that I'm shaky on "strange integers" as well — I believe your solution and use of quadratic integers is correct and valid. As to your embedded question about the use of computer applications in a proof or solution... It in no way invalidates your approach. However, there are several things to consider. If you ... 2 Unlikely that there is anything useful. The trouble is that we can take any highly abundant number, even colossally abundant, which is therefore of the form$$ 2^{e_2} 3^{e_3} 5^{e_5} \cdots p_r^2 p_{r+1}^2 \cdots p_{R}^2 \cdot p_{R+1} \cdot p_{R+2} \cdots p_N $$with e_2 \geq e_3 \geq e_5..., and simply replace p_N by the next prime, p_{N+1}. The ... 0 I wrote the following JavaScript which is designed to run in Windows Script host: isprime(n) - determine whether a number is prime via brute force method showmagic(f, txt) - writes a line to both screen and text file findmagic(f, n) - detects and writes the magic for a given prime main - tests the first 500 numbers Save the script to a JavaScript file, ... 0 This problem is extremely famous in world of Olympiad mathematics(It appeared in IMO 1990 , Problem-3) . This problem is famous, because it is also very easily solvable by use of Lifting the exponent lemma If you don't know about this lemma yet, then read this article, you will never regret. Otherwise, see this. 1 Consider the set of all multiples of 2. This is a commutative ring. 42, 66, 70, and 110 are all irreducible, and not associates of each other, and 4620=42\times110=66\times70. 3 The ring \mathbb{Z}[\sqrt{-5}] of complex numbers of the form a+b\sqrt{-5} with a,b\in \mathbb{Z} is not a UFD because 6=2\cdot 3 and 6=(1+\sqrt{-5})(1-\sqrt{5}); none of 2,3,1+\sqrt{-5},1-\sqrt{-5} are associates, so even when we make these irreducible factorizations (they actually are already), they won't be the same. 1 Maybe if we write N=\lfloor 10^{2008}\cdot 0.\overline{111}\rfloor=\dfrac{10^{2008}-1}{9}. Then we must have$$ \sqrt{N}\approx\sqrt{N+\tfrac19}=\sqrt{\frac{10^{2008}}{9}}=\frac{10^{1004}}3=10^{1004}\cdot 0.\overline{333} $$and with the derivative of the square root function f(x)=\sqrt x being f'(x)=\frac{1}{2\sqrt x} the difference between \sqrt{N} ... 2 Note that N=\frac{10^{2008}-1}{9}. One can use the Taylor series of \sqrt{x} at a=10^{2008} to solve this problem. More precisely, let f(a)=\sqrt a,$$f(x)=f(a)+(x-a)f'(a)+\frac{f''(a)}2(x-a)^2+\cdots$$Or$$\sqrt{9N}=10^{1004}-\frac{1}{2\cdot 10^{1004}}-\frac{3}{4\cdot 10^{3\cdot 1004}}+\cdots$$That means$$3\sqrt{N}=10^{1004}-5\cdot ... 1\log(1-1/x)=-1/x-O(1/x^2)$and$\pi(x)=x/\log x+O(x/\log^2x)$so their product is$-(1/x+O(1/x^2))(x/\log x+O(x/\log^2x))=-1/\log x-O(1/\log^2x).$0 Alternatively, you can note that$n = \phi \ast u$where$u = 1$for all n, and$\ast$denotes the Dirichlet product, and use the Mobius inversion formula, so$n = \phi \ast u\\ n \ast \mu = \phi \ast u \ast \mu = \phi \ast I = \phi$where$I$is the identity function, and so$\phi = n \ast \mu = \sum_{d|n} d \, \mu \left( \frac{n}{d} \right)$by ... 1 Answer: If you do not know how to use modular arithmetic, here is an interesting observation: Remainder of 6789/999 = 795 Remainder of 67890000/999 = 957 Remainder of 678900000000/999 = 579 This repeats itself for every addition of 10000. 300 digits/(3*4) = 25 Sum of the remainders = 25*(795+957+579) = 58275 Remainder of 58275/999 = 333. Two other ... 2 Similar to casting out nines to find the remainder when divided by 9, you can cast out 999's to find the remainder when divided by 999. The reason this works is the same: 1000 = 1 modulo 999, so xyz000...000 = xyz modulo 999. But instead of just adding up the digits, you have to add three-digit groups. Take for instance the number 12345678. Break this into ... 1 We have $$6789\sum_{r=0}^{74}10^{4r}$$ But$\displaystyle 10^1\equiv10,10^2\equiv100,10^3\equiv1\pmod{999}$$$\implies \sum_{r=0}^{74}10^{4r}=\sum_{r=0}^{24}(1+10+100)\pmod{999}\equiv25(111)$$ We need to find$6789\cdot25\cdot111\pmod{999}\displaystyle25\equiv-2,6789\equiv3\pmod9\implies25\cdot6789\equiv-2(3)\equiv-6\equiv3\pmod9$... 0 Partial Proof (Thanks to @hanu for providing the base proof I will use) If$m = 2^n - 1$and$m$is a Carmichael number, then$n$must be even. First, hanu's proof verified that$\lambda = n$where$\lambda$is the Carmichael function. We know that$\lambda = lcm(p_1−1,p_2−1,p_3−1,⋯,p_r−1)$. Now its very obvious that for all numbers$n > 2$,$\lambda ...

7

First, if $G$ is connected, then $G/H$ is connected (the map $G\to G/H$ defining the quotient" is faithfully flat, in particular surjective). Take a connected linear algebraic group with disconnected center, e.g., $\mathrm{SL}_2$ in characteristic zero. The quotient by the center is connected, but the center isn't connected (it's étale and non-trivial). In ...

14

Since $$2^n\equiv -n\pmod{2^n+n}$$ we deduce $$8^n = (2^n)^3 \equiv (-n)^3 \pmod {2^n+n}$$ So $2^n+n\mid 8^n+n$ if and only if $2^n+n\mid n-n^3$. For $n\geq 10$, $2^n>n^3$ so $2^n+n$ cannot divide $n^3-n=-(n-n^3)$. Clearly, if $n=0,1$, $n^3-n=0$ so $2^n+n\mid n^3-n$. So you really only need to check additionally $n=2,3,4,5,6,7,8,9$ by hand. You get ...

1

This is a special case of the Landau-Ostrowski Diophantine equation $$ay^2+by+c=dx^n$$ with $x=2$, $y=n$, $c=-1$, $b=0$, $d=1$ and $a=k$, i.e., $kn^2-1=2^n$. It has the solutions $(n,k)=(1,3),(3,1)$. This follows, for example, from Theorem $L$ in the paper On the number of solutions of the generalized Ramanujan-Nagell equation by Y. Bugeaud. In general, ...

1

We want to find those $n\in\mathbb N$ (we exclude the $0$) s.t. $$2^n=kn^2-1,$$ with integer $k$. To do so, we introduce the curves $y_1(n)=2^n$ and $y_k(n)=kn^2-1$, with real nonnegative variable $n$ and we study their intersections. We are interested in integer intersections. Let us do this. $k = 0$ We are left to solve $2^n=-1$, which has no ...

10

It converges to $0$, in fact $\frac{1}{k^7\sin{k}}$ already converges to $0$, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of $\pi$ as the borderline number $\mu$ such that $\frac{1}{k^{u-1}\sin{k}}$ converges to $0$ for $u>\mu$, and diverges for $u<\mu$. So $\frac{1}{k^7\sin{k}}$ converges because $\mu$ ...

0

NOT A PROOF. But here are my thoughts. I don't think it will converge. As k continues to get larger, it will hit closer and closer multiples of pi. For example, k=3, 31, 314, 3141, 31415, ... etc. As these values of k occur, sin(k) will get closer and closer to zero. Since sin(k) is in the denominator, the effect will cause the $k^{th}$ sequence element to ...

0

It may be hard to find a proof or a counter-example, since the lower-dimensional Pollock's conjectures are still conjectures, too.

1

First of all, if $a=b=0$, then any pair $(z_x,z_y)$ is a solution. If, only one of $a$ or $b$ equals $0$, then you have no solutions. Now we'll assume that $ab\neq 0$ You can rewrite your equation as $\frac{a}{b}=\frac{z_y}{z_x}$. This gives us a geometric intuition: we want $(z_x,z_y)$ to be a lattice point, collinear with the origin and $(b,a)$, lying ...

2

The answer is 1. The result is known as Goldbach-Euler theorem. See Wikipedia entry for "proof". For rigorous proof, you could consider sum of reciprocals of all perfect powers, $S$. Note that sum equals $$S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = ... 0 Let's see. say e have K places in the row, N men and M women. First we consider each man and each woman as a distinct object, then we see what changes if we distinguish people only by sex, i.e. if each man is considered identical to each other and same for women. We suppose K=M+N, so the people we have fill the whole row. First, we calculate K! and ... 0 Just some initial observations: Suppose m=2^n-1 and suppose m is Carmichael. If p is prime and p \mid m, then p -1\mid m-1=2(2^{n-1}-1). Since 2^{n-1}-1 is odd, we must have p \equiv 3 \pmod 4 for all p \mid m. For n \ge 2, m \equiv 3 \pmod 4. m is Carmichael and hence square free. If$$m = \prod_{i=1}^kp_i\qquad\text{for $p_i$ ...

1

$2^x+17=y^2$ leads to the three elliptic curves $y^2=u^3+17$, $y^2=2u^3+17$, and $y^2=4u^3+17$ (depending on $x$ modulo $3$). There are standard techniques for solving these, that is, for finding all the integral solutions. Indeed, you can write each one as a Mordell equation, $Y^2=U^3+k$; for example, from $y^2=4u^3+17$ we get $16y^2=64u^3+272$, so ...

-1

I get the same result as Barak, as follows: $$\sum\limits_{k\in A}\frac{1}{k-1}>\sum\limits_{k\in A}\frac{1}{k}$$ Now fix $n\in\mathbb{N}$. For each $n$ the latter sum includes (all of) $\zeta(2)-1$, $\zeta(3)-1$, ..., $\zeta(n)-1$, therefore, $$\sum\limits_{k\in A}\frac{1}{k}\sim\sum\limits_{k=2}^n\zeta(k)-(n-1)$$ and we have (see here) ...

0

Not a complete answer to your question, but if $A$ is a multi-set (where any element may appear more than once), then your series is larger than the sum of all the following series put together: $\sum\limits_{n=2}^{\infty}\frac{1}{2^n}=\frac{1}{2-1}-\frac{1}{2}=\frac{1}{2}$ $\sum\limits_{n=2}^{\infty}\frac{1}{3^n}=\frac{1}{3-1}-\frac{1}{3}=\frac{1}{6}$ ...

1

You can find the order of 10 (mod x). To do this, you just find the least k such that $10^k=1$ (mod x), $k>0$. Then $10^{r+k}=10^r$, and you know the remainder of $10^n$, $10^{n-1}$ etc so you can just add them up. Update: If you have such large modulus relative to n, then using order will not help. $ord_{2123}(10)=192$, so unless $n>192$ you won't ...

0

Partial Proof I have found a proof for numbers with an even number of factors, and a prime exponent. ($m = 2^p-1$ and $p > 2$) First note that $2^p-1 \equiv 3 \mod 4$ for $p > 1$. Next note the following table for $a*b \mod 4$: $$\begin{array}{c|lcr} b & a = 0 & a = 1 & a = 2 & a = 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 ... 0 We know that we can find integers k,\ell such that$$a = b + kmb = 1+ \ell mn$$Hence,$$a = 1+\ell mn+km = 1 + (\ell n+k)m$$So if we want a \equiv 1 \pmod{mn} then we need \ell n+k \equiv k \equiv 0 \pmod{n}. Hence$$a \equiv b \pmod{mn}

3

Yes, there is. Since $X=\frac{10^N-1}{9}$ and we can find the inverse of $9$ in $(\mathbb{Z}/M\mathbb{Z})^*$ in $O(\log M)$ time with the extended Euclidean algorthm, we just need to find $10^N\pmod{M}$. This can be done in $O(\log N)$ time with the repeated squaring algorithm. If you know the factorization of $M$ or a large portion of primes dividing ...

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