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1

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}


0

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


1

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


0

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 ...


0

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? :)


1

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 ...


0

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


2

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 ...


0

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


0

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.


1

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


2

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


1

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 characteristic $p$, 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: $\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}$ $v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}x$ $x_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}$ $y_3 = -(\lambda x + v)$ (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 $w_1$ in 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 factorizations $$2012=1\cdot2012,2\cdot1006,4\cdot503$$ $$2013=1\cdot2013,3\cdot671,33\cdot61,11\cdot183.$$ 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 parameters $k,s,t$ - 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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