# Tag Info

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This is Mathematica code. I believe that it will work on Wolfram Alpha. m = 11; Select[Range[m], Mod[2^#, m] == # &] {7} For large values of $m$ this will be faster m = 1234567; Select[Range[m], PowerMod[2, #, m] == # &] {313692}

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Whether the limit does exists is not known now,but the gap between two consective prime numbers can be arbitrarily large.

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Add $1$ to each of $x_1$ and $x_2$, and divide throughout by $2$ to get $y_1+\ldots+y_p=\frac{n}{2}+1$, all of the $y_i$'s non-negative. Then number of integer solutions which boils down to star and bars

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Note that if $r\equiv x^2\equiv y^2 \mod p$ then $x^2-y^2=(x+y)(x-y)\equiv 0 \mod p$ Since $p$ is prime, one of the factors $x+y, x-y$ must be divisible by $p$, and this is equivalent to $x\equiv \pm y \mod p$. Apart from the residue zero, and the case $p=2$ where we have $1=-1$, this gives precisely two square roots for each quadratic residue. The number ...

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Here I have a solution just using Pythagorean triples. we know that any solution of $$x^2+y^2=z^2$$ can be written as $$x=2ab,\,\,\,\,\, y=a^2-b^2,\,\,\,\,\ z=a^2+b^2.$$ Therefore for your equation, we can choose $$w=4abc,\,\,\,\,\, x=2(a^2-b^2)c,\,\,\,\,\ y=(a^2+b^2)^2-c^2,\,\,\,\,\,z=(a^2+b^2)^2+c^2.$$ Isn't it interesting? :)

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Expanding David's answer, for any $p>2$ the map $\phi:\mathbb{F}_p^*\to\mathbb{F}_p^*$ defined by $\phi(x)=x^2$ sends $y$ and $-y$ into the same element, hence the number of quadratic residues is $\leq\frac{p-1}{2}$. On the other hand, $\mathbb{F}_p^*$ is a cyclic group generated by $g$ with $o(g)=p-1$, hence all the "even powers" of $g$ are quadratic ...

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Modulo a prime (except for $2$), exactly half the non-zero residues are quadratic residues.

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I don't think you're going to find any characterization that's simpler than your definition itself. The boundary between convergence and divergence is very delicate. For example, if $x_n = n( \log n )(\log\log n) (\log\log\log n)^\alpha$, then $S_*$ converges regardless of the value of $\alpha$, but $S^*$ converges when $\alpha>1$ and diverges when ...

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I think we can apply the trick which Gauss applied to sum up first 100 numbers. Essential idea: (p-1)! % p = (1 * 2 * 3 * 4 * .... * (p-1) % p = {(1 * (p-1)) * (2 * (p-2)) * (3 * (p-3)) ...) % p --> rearranging terms = { (1 * -1) * (2 * -2) * (3 * -3).... ) --> taking % p inside

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The long Weierstrass form $y^2+xy+y=x^3$ is transformed into the short Weierstrass form, namely to $$y^2=x^3+621x+9774.$$ The formulas for the necessary substitutions are given here.

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$$(n^2+3n)(n^2+3n+1)\lt n(n+1)(n+2)(n+3)\lt(n^2+3n+1)(n^2+3n+2)$$

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$$n(n+1)(n+2)(n+3) = (n^2 + 3n)(n^2 + 3n +2)$$ If $n^2 + 3n \geq k$, then $(n^2 + 3n)(n^2 + 3n +2) > k(k+1)$ If $n^2 + 3n <k$, then $(n^2 + 3n)(n^2 + 3n +2) < k(k+1)$

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You can write using partial summation formula, \displaystyle \begin{align} \sum\limits_{p \le x} \dfrac{\log^2 p}{p} = \sum\limits_{p \le x} \log p\dfrac{\log p}{p} &= \log x\sum\limits_{p \le x} \dfrac{\log p}{p} - \int_2^{x} \frac{\sum\limits_{p \le t} \frac{\log p}{p}}{t}\,dt \\&= \log x(\log x + \mathcal{O}(1)) - \int_2^{x} \frac{\log t + ... 1 In characteristicp$, where$p$is odd, $$(1+X)^{p-1} = \frac{(1+X)^p}{1+X} = \frac{1+X^p}{1+X} = \frac{1-(-X)^p}{1-(-X)} = 1 -X + X^2 - \dots +X^{p-1}.$$ 2 More precisely,$M_p$is a pseudoprime to the base$2$. To show this we show that $$2^{M_p-1}\equiv 1\pmod{M_p}.$$ By Fermat's Theorem we have$2^{p-1}\equiv 1\pmod{p}$. Thus$2^{p-1}=1+kp$for some integer$k$, and therefore$M_p-1=2kp$. Thus $$2^{M_p-1}=(2^p)^{2k}=(1+M_p)^{2k}\equiv 1\pmod{M_p}.$$ 1 $$\binom{p-1}k=\frac{(p-1)\cdots(p-k)}{k!}=\prod_{r=1}^k\frac{p-r}r$$ Now$p-r\equiv-r\pmod p\implies\dfrac{p-r}r\equiv-1$1 You can prove it by induction on$k$. If$ k=1\to\displaystyle \binom{p-1}{k} = p-1$that$p-1 \equiv -1 \mod p$. For$k= n +1$use this$\displaystyle \binom{m}{n+1} =\displaystyle \binom{m}{n} +\displaystyle \binom{m}{n+1}$6 Let$a=\binom{p-1}{k}$. Then $$a k!=(p-1)(p-2)(p-3)\cdots (p-k).$$ The$i$-th term on the right-hand side is congruent to$-i$modulo$p$. Thus $$ak!\equiv (-1)^k k!\pmod{p}.$$ Now since$k!$is not divisible by$p$we can cancel. 0 Aspects of your question have been covered here about Lebesgue's Identity, lebesgue's identity and a general approach to$x_1^2+x_2^2+\dots+x_n^2 = z^2$, Diophatine equation$x^2+y^2+z^2=t^2$0$s=\dfrac{a(a+d)(a+2d)...(a+10d)}{\gcd\{a,(a+d),...,(a+10d)\}(a+10d)}=\dfrac{a(a+d)(a+2d)...(a+9d)}{\gcd\{a,(a+d),...,(a+10d)\}}$Clearly the A.P. is increasing, so if we consider$D$to be the$\gcd$for brevity, we must have$D\leq a$. Since we are looking for the smallest$s$it suffices to consider$D=a$. Now$D|(a+d)\implies a|d\implies d=at$for ... 0 In General it is possible in various ways to write the solution to this equation: $$x^2+y^2+z^2=q^2$$ I like such kind. $$x=2a^2s^2-2abs^2\pm{2apbs}$$ $$y=2a^2s^2+2abs^2\pm2apbs$$ $$z=p^2b^2-a^2s^2+s^2b^2\pm2apbs$$ $$q=p^2b^2+3a^2s^2+s^2b^2\pm2apbs$$ If you want you can write infinitely many formulas are not the problem. You can even choose a ... 0 Hints: Any integer can be written one of$m$numbers:$km,km+1,\cdots, km+(m-1)$2 Let the numbers be$b_r=a+r, 0\le r\le m-1$Existence: We can apply Pigeonhole Principle to prove the existence by contradiction. Let none of them is divisible by$m,$so they can leave$m-1$distinct remainder$(r)$s namely,$1\le r\le m-1$But, as there are$m$numbers, so at least tow of them leave the same remainders. Let$b_u,b_v$leave the same ... 1 This is very probably an open problem. It is closely related to Pillai's conjecture which is a generalization of Catalan's conjecture. The former looks for solutions in the integers to $$a^n - b^m = c$$ Catalan's conjecture is that$a=m=3$and$b=n=2$is the only solution for$c=1$. It was proven in April 2002 by Pedra Mihăilescu. EDIT: @RobertIsrael ... 2 See OEIS sequence A074981 and references there.$10$does have a solution as$13^3-3^7$, but apparently no solutions are known for$6$and$14$. 1 We have: $$\tag 1 y^2 = x^3 + ax + b$$ To add a point, we have:$\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}xx_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}y_3 = -(\lambda x + v)$(Calculate this out) Next, we can substitute$(1)$... 2 Notice$2$is a primitive root modulo$11$. In other words, any number between$1$and$10$is congruent modulo$11$to a unique power$2^k$for$1 \leq k \leq 10$. Notice$10 \equiv -1$, so on your first congruence you're really trying to solve$x^5 \equiv -1 \pmod {11}$. Let's suppose$2^k$is a solution for$x$(so to find$x$, you'll have to solve ... 1 The functions$f_1(x)=\sin(x)$and$f_2(x)=\cos(x)$are transcendental but satisfy$P(x,f_1(x),f_2(x))=0$for$P(x,y_1,y_2)=y_1^2+y_2^2-1$. The concept you have described is algebraic dependence. 0 Now try a substitution$x=u+k$with$k$a constant, multiply out and choose$k$so there is no$u^2$term. With$k=-1/12$the cubic in$x$then in terms of$u$is $$u^3+\frac{23}{48}u+\frac{181}{864},$$ if my calculations are OK. Anyway that's the idea to finish from where you are. 3$c = 2859545$factors as$5 \times 13 \times 29 \times 37 \times 41$. Each of these primes is congruent to$1$mod$4, so they factor over the Gaussian integers: $$5 = \left( 1+2\,i \right) \left( 1-2\,i \right) , 13 = \left( -3+2\,i \right) \left( -3-2\,i \right) , 29 = \left( 5-2\,i \right) \left( 5+2\, i \right) , \left( 1+6\,i \right) \left( ... 2 I have found a significant number of polynomials of this type with T\ge 24 but nothing with T>40. The prime k-tuples conjecture suggests that there should be examples with T arbitrarily large since 2n^2 and n^2+n omit some residue classes for every prime. For example, for n=0,1,\ldots,9 the differences n^2+n-(0^2+0) are ... 3 Factor 2859545 = 5 \cdot 13 \cdot 29 \cdot 37 \cdot 41. All these primes are of the form 4k+1 and so can be expressed as sums of two squares in essentially one way. You can combine the solutions for each prime into several different solutions for 2859545 using Brahmagupta's identity. 0 According to Maple,$$\eqalign{m^e &\equiv 14178120117339266261904109624890227407523673482786024455164108238811626728989472\cr &1928886528279400665340976615948053755946789302972467196231829204061441862114262 \cr &11704026304276898946101664519229545142128929712389990917298990673103791915511466 \cr ... 2 Do you know the group law (of an elliptic curve)? Assuming we're on a field of characteristic\;\neq 2,3\;$,we can define: $$t:=\frac{3\cdot 2^2}{2\cdot 1}=6$$ $$x_1:=t^2-2\cdot2=32\\y_1:=1+6(32-2)=181$$ and we get a new solution$\;(32\,,\,\,181)\;$0 Revised in response to comments. Let’s look at the case in which you can put the weights on both pans. Suppose that you have weights$w_0,w_1,w_2,\ldots\,$, where$w_0<w_1<w_2<\ldots\;$. You want to weigh an$n$-pound object. Let’s say that the scales balance when you have$w_3$and$w_0$in one pan, and the object and$w_1in the other pan. This ... 0 Solving the first two equations simultaneously, you get x = 24(mod 110). Solving this result and the third equation simultaneously, you get x = 134(mod 330). 1 Mertens' third theorem is just the exponentiated version of the second theorem (without the bounds that Mertens proved for his second theorem): \begin{align} -\ln\Biggl(\ln n\prod_{p\leqslant n}\biggl(1 - \frac{1}{p}\biggr)\Biggr) &= -\ln \ln n - \sum_{p\leqslant n} \ln \biggl(1 - \frac{1}{p}\biggr)\\ &= \Biggl(\sum_{p\leqslant n}\frac{1}{p} - \ln ... 0 (Too long for a comment.) I simplified your expression and found they are ternary quadratic forms. (Why didn't you just simplify them? Maple and Mathematica can do it easily.) So, $$R^n+Q^n+T^n = X^n+Y^n+Z^n,\quad for\; n =2,4\tag{1}$$ \begin{align}R =& -2 k^2 - 2 k s + s^2 + 3 k t - t^2\\ Q =&\; k^2 - 2 k s - 2 s^2 + 3 s t - t^2\\ T =&\; ... 2 Start from 2012=2^2\cdot503 and 2013=3\cdot11\cdot61, giving you all possible factorizations2012=1\cdot2012,2\cdot1006,4\cdot5032013=1\cdot2013,3\cdot671,33\cdot61,11\cdot183.$$This gives you all possible assignments of a-b,c-a,b-c,d-c. From there, deduce b-d and compute the new ratios. There will be 48 solutions unless some are equal or ... 0 Since you have four unknowns, you are going to have infinitely many solutions. If your job is to find just one 4-tuple that satisfies your first equation, I'd just pick numbers that make it very easy. For example, choose c=0, d = 1, b=2. If you use these particular numbers, you will need to use the quadratic formula to solve for a at the end. 0 With a=2013,b=1,c=2014 and d=2013,$$\frac{(2013-1)(2014-2013)}{(1-2014)(2013-2014)}=\frac{2012}{2013}$$1 Note that$$\overline{x}_i=17i-3$$is a solution for i=1,\cdots, 588. But,$$x_i=17i+3$$is a solution for i=0,\cdots, 588. Thus, x_0=3 is a solution that is not computed in the book. That is, your answer is correct. 0 Your inequality is equivalent to$$-\underset{p\leq x}{\sum}\log\left(p^{N+1}-1\right)>\log\left(0.2\right)-2\log\left(\log\left(x\right)\right).$$Now we have, for partial summation and Prime Number Theorem, that exists c_{1},c_{2}>0 such that$$-\log\left(0.2\right)-\underset{p\leq x}{\sum}\log\left(p^{N+1}-1\right)<-\underset{p\leq ... 1 I think not to introduce additional equations, and directly solve the system of equations. \left\{\begin{aligned}&R^2+Q^2+T^2=X^2+Y^2+Z^2\\&R^4+Q^4+T^4=X^4+Y^4+Z^4\end{aligned}\right. Using integer parametersk,s,t$- Will make a replacement. $$a=3(k+s-t)^2+k(k-t)$$ $$b=3(k+s-t)^2+s(s-t)$$ $$c=3(k+s-t)^2-t^2+(k+s)t-2ks$$ ... 4 Consider the limit $$\lim_{n\to \infty} \frac{E_n}{n!}\left(\frac{\pi}{2}\right)^n$$ Using$E_{n} \sim \frac{(-1)^{(n-1)/2} 4^{n+1}}{n+1}B_{n+1}$together with$B_{2n} \sim (-1)^{n+1}4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$and Stirlings approximation we get $$\lim_{n\to \infty} \frac{E_{2n-1}}{(2n-1)!}\left(\frac{\pi}{2}\right)^{2n-1} = ... 3 By Abel's Theorem, if the power series converges, then it agrees with the limit of the function. This argues against -2/\pi. Instead, either the power series converges to 0 (by the limit Michael & Ewan give in the other answer) or it does not converge. Some messing around numerically: It looks like the terms of the sum (not the sum itself) ... 4 Here’s a proof using a solution I found here. Write n in the form 8^m\cdot s, for an integer m\ge0 and with s not divisible by 8. This can always be done. Note that an integer s that is not a multiple of 8 can be written in one of the following three forms: 2k+1 (if s is odd), 4k+2 (if s is even, but not a multiple of 4), or 8k+4 ... 2 It's as simple as: If n \nmid 3, then n \equiv 1 or 2 \pmod 3. So n^2 \equiv 1^2 = 1 or 2^2 = 4 \equiv 1 \pmod 3, i.e. n^2 \equiv 1 \pmod 3, giving the required result. 5 My friend put this as an MAA Monthly problem, years ago. The comparison is that there are infinitely many numbers that have no expression as x^2 + y^2 + z^9. This simple result defeated an existing conjecture; we sent it early to Robert C. Vaughan, so it made it into the second edition of his book The Hardy-Littlewood Method. It is likely that every number ... 2 Note that$$(3k+2)^2=9k^2+12k+4=3m+4=3m+3+1=3(m+1)+1=3l+1$$Or rewrite your integers as$$3k-1 , 3k, 3k+1$$and then$$(3k-1)^2=9k^2-6k+1=3(9k^2-6k)+1=3p+1$\$ I hope you can handle the rest!

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