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How to check for solutions of non-linear systems over the integers modulo m?

Let's see, lemma, given a prime $p$ and a target $t,$ there always exist $x,y$ such that $x^2 + y^2 \equiv t \pmod p.$ This one uses just a counting/pigeonhole argument This can always be done for ...
Will Jagy's user avatar
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1 vote
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A question about integrality in polynomial rings

Pick an $\alpha \in B$, and consider $B'$ the subring of $B$ generated by the $x_i$'s and the coefficients of $\alpha$ (denoted $c_1,...,c_m \in \overline{k}$ in no particular order). Then $\alpha$ is ...
Fançois Gatine's user avatar
7 votes
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Approximating the Prime Counting Function as $\pi(x) \approx \frac{x^2}{\ln\left(\Gamma(x+1)\right)}$

Since $\log \Gamma(x+1) = x\log x - x + O(\log x)$, \begin{align*} \frac{x^2}{\log \Gamma(x+1)} &= \frac x{\log x - 1 + O(x^{-1}\log x)} \\ &= \frac x{\log x} \biggl(1 - \frac1{\log x} + O(x^{-...
Greg Martin's user avatar
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14 votes

What's the easiest way to prove the Riemann Zeta function has any zeros at all on the critical line?

The easiest way is to refer to the Hardy $Z$ function, $$ Z(t) = e^{i \theta(t)} \zeta(1/2 + it), $$ where $$ e^{i \theta(t)} = \pi^{-i t / 2} \frac{\Gamma(1/4 + it/2)}{\lvert \Gamma(1/4 + it/2) \...
davidlowryduda's user avatar
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How to prove that$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O (x) $

Most answers here assume $$\psi(x) - \vartheta(x) = O(x^{1/2}),$$ which I believe is only proven later in Apostol's Analytic Number Theory book. Prior to the exercise mentioned by OP, the author only ...
huh's user avatar
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Doubt in proof of theorem 8.20 of Apostol Modular functions and Dirichlet series in number theory

This is a famous result that the partial sums $\zeta_{n}(s)$ has zeros even for $\sigma > 1$, and it must be true since many have already read the argument and bought it. I guess the author of this ...
sozinhozinho's user avatar
0 votes

The "Euler Product formula" for general multiplicative functions

A lot can be deduced from the following property: \begin{equation} \left(1-\frac{1}{x}\right)\left(1-\frac{1}{x-1}\right)\cdots \left(1-\frac{1}{x-k}\right)=1-\frac{k+1}{x} \end{equation} Now, we ...
Zima's user avatar
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0 votes

The average number of prime factors of a number less than or equal to $N$

Yes, indeed. For each positive integer $m$, we have \begin{equation} \sum_{n \le x} \omega(n)= x\log(\log x)+Mx+\sum_{k=1}^{m}C_k \frac{x}{(\log x)^k}+O\bigg(\frac{x}{(\log x)^{m+1}}\bigg) \hspace{...
AfterMath's user avatar
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1 vote
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The integral basis of $\mathbb{Q}(\zeta+\zeta^{-1})$

First, a priori you want to show that every element of $\mathcal{O}_K$ is of the form $\sum_{i=0}^{\frac{p-3}{2}} t_i (\zeta+\zeta^{-1})^i$ with $t_i\in\mathbb{Z}$, and not of the form $t_0+ \sum_{i=1}...
Enguerrand Moulinier's user avatar
2 votes

Finiteness of a generalization of the class number of a number field

It is true that the number of isomorphism classes of full $\mathcal{O}$-modules in $\mathcal{K}^{\textbf{n}}$ is finite. The result is a special case of the Jordan-Zassenhaus Theorem. For future ...
Ben Marlin's user avatar
1 vote
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Discriminant of numbers

It is a standard calculation in number theory that $\text{disc}(x^n-1) = (-1)^{n(n+1)/2 + 1}n^n$, and this implies that $\text{disc}([n]_q) = (-1)^{n(n+1)/2 + 1}n^{n-2}$. Let me prove both these ...
SomeCallMeTim's user avatar
3 votes

Fermat's last Theorem and elliptic curve cryptography

There is no relationship between the two. FLT is about homogeneous equations of the form $x^n+y^n=z^n$ over the integers and states that no integer solution exists when $n\geq 3.$ Yes it is stated in ...
kodlu's user avatar
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0 votes

$H(x)$ approximates $\pi(x)$ pretty well. But what are the drawbacks, when compared with Riemann's $R(x)$?

When $z$ is small, there is $$ e^z=1+z+O(z^2), $$ so we have $$ {e^z-1\over e^z+1}={z+O(z^2)\over2+O(z)}=\frac z2+O(z^2). $$ This indicates that your $f(x)$ (plug in $z=2/\log n$) is $$ f(x)\approx\...
TravorLZH's user avatar
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1 vote

Fermat's last Theorem and elliptic curve cryptography

I don't think there is any connection and the temporal link is weak. First, Fermat's last theorem was proven in 1994. Second, elliptic curve cryptography was proposed in 1985, so about 10 years before ...
quarague's user avatar
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4 votes
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2-adic valuations of $k\cdot 3^n-1$

The best way to understand the situation, for odd $k$ (even $k$ is trivial) is to decompose your expression, to wit: $(k\cdot3^n)-1=k[(3^n-1)-(k^{-1}-1)]$ where $k^{-1}$ is the reciprocal of $k$ in $2$...
Oscar Lanzi's user avatar
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2 votes

A simple question about bounding a sum

First, the condition $\|x\|_2>\gamma^{-1}$ leads to $\frac{1}{\|\gamma x\|_2} <1$, which implies that $\frac{\gamma^2}{\|\gamma x\|_2^2}$ dominates the other terms in the sum. Hence, we only ...
alex440's user avatar
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2 votes
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Ramification for a Galois extension of number fields

You can find a proof in these lecture notes: https://websites.math.leidenuniv.nl/algebra/ant.pdf See theorem 4.14, and also theorem 4.10 if you need the relation between the discriminants of $f$ and $...
Servaes's user avatar
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2 votes

Fibonacci and Lucas numbers related identities

By definition $$ \sum_{i=1}^n H_i = \sum_{i=1}^n L_i +m \sum_{i=1}^n F_i $$ where $L_{i\ge 1}=1,3,4,7,\ldots$ and $F_{i\ge 1} = 1,1,2,3,5,\ldots$. We have a well-known partial sum, see https://oeis....
R. J. Mathar's user avatar
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3 votes
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Detailed Proof of Proposition 2.12 a) from Diamond, Darmon, Taylor, "Fermat's Last Theorem."

You need to show that for all $\sigma \in G_p$, all $n \ge 1$, and all $\ell^n$-th roots $Q$ of $q$ in $\overline{\mathbf{Q}}_p$, we have $\frac{\sigma(Q)}{Q} \in \mu_{\ell^\infty}$. But $\sigma$ is a ...
David Loeffler's user avatar
3 votes
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Is $\frac{a^{p}-1}{a-1}$ ($p$ prime) square-free?

Converting my comments into an answer: This is false for every prime $p$, and the smallest counterexample is $\boxed{ \frac{3^2 - 1}{3 - 1} = 2^2 }$. For any prime $p$, counterexamples can be ...
Qiaochu Yuan's user avatar
0 votes

Show that no set of nine consecutive integers can be partitioned into two sets where the products of all elements of two sets are equal.

OP observes that the sequence must be composed of integers that have no prime factors greater than $7$. We also know that if the two subsets have an equal product, then the product of all members of ...
Keith Backman's user avatar
1 vote

Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

I think this should always be true. We have a natural map $\lim\limits_{\rightarrow}(A_i[2])\rightarrow(\lim\limits_{\rightarrow}A_i)[2]$ induced by the inclusions $A_i[2]\hookrightarrow A_i$. We can ...
LittleBear's user avatar
2 votes
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Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

This is true if the colimit is filtered. Then $M[2] = \textrm{ker}(M \stackrel{2}{\to} M )$ is the kernel of a canonical map, and kernels commute with filtered colimits for abelian groups. In other ...
Andrea Marino's user avatar
2 votes

How to find the $\zeta$ representation of a $L$-series

Another profitable way to think about these problems is via the Euler product, assuming the coefficients of the Dirichlet series are multiplicative. In the example you gave, we have $$\sum_{n = 1}^\...
CJ Dowd's user avatar
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2 votes

Prove the binary icosahedral group is isomorphic to ${\rm SL}(2,5)$

Your strategy looks fine to me, let's try to make it work. First, working $\bmod \sqrt{5}$ we can invert $2$ and $\tau \equiv 3 \bmod \sqrt{5}$, so we get $\zeta \equiv -1 + i + 3j \bmod \sqrt{5}$. So ...
Qiaochu Yuan's user avatar
5 votes

Is A276175 integer-only?

Following up on a comment by David E. Speyer (indicating that the $b_n$ sequence arises from the cluster algebra $\mathcal{A}(Q)$ associated to a certain quiver $Q$), I would like to mention that the $...
Antoine de Saint Germain's user avatar
0 votes
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Galois Group of $\mathbb Q$ Surjects onto Certain Cyclotomic Extension

This ultimately reduces to finite Galois theory using topology. Let me explain. Claim 1: let $G$ be a compact group, and $H$ be a profinite group and $f: G \rightarrow H$ be a continuous homomorphism ...
Aphelli's user avatar
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Is this polynomial irreducible over the rationals?

We can use a bit of Galois theory. Let $\zeta_n = e^{2\pi i/n}$, so $2\cos \frac{2\pi k}{n} = \zeta_n^k + \zeta_n^{-k}$. Since $T_n(\cos(2\pi k/n)) = \cos(2\pi k) = 1$, we have $\cos(2\pi k/n)$ is a ...
arkeet's user avatar
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0 votes

Number theory problem involving linear congruences

To answer as the OP says "involving linear congruences," suppose the three integers are $a$, $b$, and $c$. Then, by the Division Algorithm, using $r$ as the remainder: \begin{align} 3a &...
k endres's user avatar
2 votes
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Norm of a simple extension

As commented by @Daniel5803, the mistake was in the calculation of the last column of the matrix, because $\phi(a^{n-1})=xa^{n-1}-a^n$ but $a^n$ is not in your basis. Let us first prove the special ...
Anne Bauval's user avatar
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0 votes

Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$ and $a+b=c+d$

We cut to the chase. Given $a+b=c+d\not=0\tag{1}$ $a^3+b^3=c^3+d^3\not=0\tag{2}$ Divide (2) by (1) to get $a^2-ab+b^2=c^2-cd+d^2\tag{3}$ but also square (1) to get $a^2+2ab+b^2=c^2+2cd+d^2\tag{4}$ ...
Oscar Lanzi's user avatar
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1 vote
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If the analytic rank is one then the sign in the functional equation is -1?

This is something you can definitely use without a reference. The functional equation is $L(f,s)=\epsilon L(f,2-s)$ for some sign $\epsilon$. Let $F(z)=L(f,1+z)$, then $F(z)=\epsilon F(-z)$. So if $\...
Aphelli's user avatar
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1 vote

How can I prove my conjecture for the coefficients in $t(x)=\log(1+\exp(x)) $?

The question posed by Gottfried Helms at How can I prove my conjecture for the coefficients in $t(x)=\log(1+\exp(x)) $? is equivalent to the known Maclaurin power series exapsnion \begin{equation}\...
qifeng618's user avatar
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7 votes
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How to show isomorphism between an adele ring on a number field and its Pontryagin dual?

(This should probably just be a comment, but I don't have enough reputation yet). One can reduce this problem to proving that local fields are self-dual by theorem 5.4, which states that the ...
Daniel5803's user avatar
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Calculation of Hilbert symbol over $\Bbb Q_p$ where $p$ is an odd prime

If you read that section it says "over the $p$-adics..." so $u, v$ are not meant to be integers but $p$-adic integers. If $a = \frac{1}{2}$ and $p = 3$ then $\alpha = 0$ and $u = \frac{1}{2}$...
Qiaochu Yuan's user avatar
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Question on a Quora Proof of Fermat's Last Theorem for Exponent 3

I believe I have concocted an answer to my question. Suppose instead, without loss of generality, that $\lambda$ divides $x$. Then write $$y^3 + u(-z)^3 = (-x)^3$$ By an exercise from the post, this ...
Johnny Apple's user avatar
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1 vote
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The equation $x^3=u$ has a solution in $\mathbb{Z}_p^{\times}$ if and only if $x^3 \equiv u \mod p$ has a solution.

First, $x^{3} = p$ implies $v_p(x^3) = v_p(u) = 0 \iff v_p(x) = 0$ so $x \in \Bbb Z_{p}^\times$. So we can write $x = a_0 + a_1p + a_2p^2+\cdots$, where $a_{i}$ is an integer between $0$ and $p-1$. ...
Afntu's user avatar
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1 vote
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Upper bound on number of divisors of $x^2 -1$

If we define $$f(n):=\frac{\frac{\ln(\sigma_0(n^2-1))}{\ln(n^2-1)}\cdot \ln(ln(x^2-1))}{1.5379 \ln(2)}$$ the above inequality transofrms for an integer $n>1$ to $f(n)<1$. In this sense , the ...
Peter's user avatar
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1 vote

Question on a Quora Proof of Fermat's Last Theorem for Exponent 3

The proofs I know start as follows. Let $(x,y,z)$ be a nontrivial integer solution to $x^3+y^3=z^3$ with coprime $x,y,z$. Then we have $3\mid xyz$, because otherwise $x^3+y^3\equiv -2,0,2\bmod 9$ and $...
Dietrich Burde's user avatar
0 votes

$\frac{1}{8}x^2+\frac{23}{9}y^2=z^2$ has non-trivial solutions $\in \mathbb{Q}$

and all integer solutions to $2x^2 + 23 y^2 = z^2$ with $\gcd(x,y,z) = 1$ and $x,y,z \geq 0$ come from $$ x = |u^2 + 10 uv + 2 v^2| \, , \; \; y = |u^2 - 2 v^2| \, , \; \; z = 5 u^2 + 4 uv + 10 ...
Will Jagy's user avatar
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1 vote

Coefficient of $n$th cyclotomic polynomial equals $-\mu(n)$

This is equivalent to proving the sum of primitive $n$th roots equals $\mu(n),$ which we can prove with principle of exclusion-inclusion. Let $\text{ord}(\omega)$ denote the order of $\omega.$ First, ...
Display name's user avatar
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$\frac{1}{8}x^2+\frac{23}{9}y^2=z^2$ has non-trivial solutions $\in \mathbb{Q}$

You want to show that, using the Hilbert symbol, that $$ \left( \tfrac{1}{8}, \tfrac{23}{9} \right)_p = +1 $$ for all primes $p$, including $p=\infty$. The case when $p=\infty$ is trivial since you ...
Nicolas Bourbaki's user avatar
4 votes
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$\frac{1}{8}x^2+\frac{23}{9}y^2=z^2$ has non-trivial solutions $\in \mathbb{Q}$

I'm assuming there is a typo and you meant 23 instead of 24 for the coefficient of $y^2$. $(x,y,z)$ is a solution to $\frac18 x^2 + \frac{23}9y^2=z^2$ if and only if $(\frac x4,\frac y3,z)$ is a ...
Yaneda's user avatar
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2 votes
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When is $x \in \mathbb{Q}_p$ a square?

If $a \in \mathbb{Q}_p^\times/\mathbb{Q}_p^{\times 2}$, then $a$ is not a square in $\mathbb{Q}_p$. I'm not sure if there is a typo in the above sentence of yours, since it is not only not true, but ...
X-Rui's user avatar
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1 vote
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Summation of n-simplex numbers

https://oeis.org/A356037 tabulates "Conjecturally, the smallest number $m$ such that every natural number is a sum of at most $m$ $n$-simplex numbers." The table doesn't go very far: just $1,...
Gerry Myerson's user avatar
2 votes
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Lemma about $n$-adic numbers

If the element $x \in \mathbb{Z}_{mn}$ is given by the numbers $x_k$, then those same numbers $x_k$ also define an element $x' \in \mathbb{Z}_n$ and similarly an element $x'' \in \mathbb{Z}_m$. Try to ...
Zufallskonstante's user avatar
0 votes

How to determine $\Gamma_{15}$?

The Chinese remainder theorem states that the ring homomorphism $\mathbb{Z}/mn\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ that maps $x + mn\mathbb{Z}$ to $(x + m\mathbb{Z},x + ...
arkeet's user avatar
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1 vote
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Factoring in $\mathbb F[x,y]$.

Over $\mathbb{F}_2[x,y]$, we might not always be able to write an ideal as a product of ideals. But what is possible is to write an ideal as an intesection of primary ideals(https://en.wikipedia.org/...
Guster's user avatar
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0 votes

Half-integer weight modular forms mod $p$; theta cycles

Choi and Lee’s papers may be good starting points: https://doi.org/10.1007/s11139-019-00154-z and https://doi.org/10.1515/math-2022-0512
Tim Huber's user avatar
2 votes
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Pairs of infinite subsets of natural numbers that intersect on an infinite set give the same property to triples?

Question 1 Counterexample: divide $\mathbb{N}$ into three infinite subsets $A_1$, $A_2$, $A_3$ and let $$\mathcal{A} = \{ A_1 \cup A_2, A_2 \cup A_3, A_3 \cup A_1 \} \cup \{ \{ m \in \mathbb{N} : m \...
Adayah's user avatar
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