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2 votes
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A function asymptotical equivalent with the prime counting function?

This is closely connected to Schinzel's Hypothesis H and the Bateman-Horn conjecture. A slightly different way to phrase this question would be to consider the two polynomials $(x, x^2 + 4)$, and look ...
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Tate gamma factor as a principal value integral

For the second question, let $\varpi$ be a uniformizer, and set $\mathfrak p = \varpi \mathcal O_F$. Let's define the conductor of $\chi$ to be $f$ if $\chi$ is trivial on $1+\mathfrak p^f$, but not ...
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Change of fractional numbers from one base to another

I'll assume you know how to write a fraction $A/B$ in decimal form, i.e. $A/B=(0.d_1d_2...)_{10}$. Then, given that $x=(0.a_1a_2...)_b=\sum_{i=1}^\infty a_i/b^i,$ just write each successive term in ...
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1 vote

Some questions in the proof of Analytic Large Sieve

I should preface by saying that I didn't read the various linked pages. The first inequality $$ |\psi |^2 = \sum_n| \sum_r a_n b_r e(n \alpha_r)|^2 \leq \sum_n |a_n|^2 \sum_n| \sum_r b_r e( n \alpha_r)...
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1 vote

Change of fractional numbers from one base to another

If we see a fraction can be represented in base 10 as $\frac{a_1}{10}+\frac{a_2}{10}+\frac{a_3}{100}....+\frac{a_n}{10^n}$ . Similarly in any base $r$ , a fraction can be represented as $\frac{a_1}{r}+...
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4 votes
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For every prime p there is a sum of squares congruent to -1 mod p

It is well known that there are exactly $\frac{p+1}{2}$ quadratic residues modulo $p$ (including $0$) for any odd prime $p$. Therefore, $a^2$ can take $\frac{p+1}{2}$ distinct values modulo $p.$ The ...
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2 votes

Why does this cycle of 44 show up in the Collatz Conjecture?

$f(x)$ for $b=6$ (bellow in Cartesian coordinates for $n=27$) is almost equivalent to the fractional part of $log_6(x)$: $g(x)=\{log_6(x)\}$ (second graph bellow) which also measure the distance to ...
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  • 2,074
0 votes

Estimating reducible monic polynomials of degree n with integer coefficients of height of atmost N

Here is a sketch of a proof: If a polynomial is reducible, then it must be reducible mod $p$ for every prime $p$. The number of degree $n$ polynomial mod $p$ that is irreducible is (Estimating ...
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Show: $\text{lcm}(a,a+p)=\text{lcm}(b,b+p), p \;\text{prime}\implies a=b$

Answer after reading the comments: WLOG $p \mid a, p \nmid b$. This implies that $pk=a, k \in \mathbb{Z}$. Then, $lcm(pk, pk+p)=lcm(pk,p(k+1))=pk(k+1)$, and $lcm(b,b+p)=b(b+p)$ So it suffices to prove ...
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2 votes

Collatz conjecture but with $n^2-1$ instead of $3n+1.$ Does the sequence starting with $13$ go to infinity?

The set of solutions can be written as table, segmented by e'th powers of $2$ : ...
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1 vote

n divides $m+1, m^m+1, m^{m^m}+1,...$

Answer after reading the comments: Let $m=2n-1$ (odd). Then the first term of the sequence works for every $n$: $n \mid m+1=2n$. To prove that it works for all the sequences' terms, we will use the ...
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2 votes

Prove that either $m$ divides $n$ or $n$ divides $m$ given that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$?

$\newcommand{\l}{\operatorname{lcm}}$ There are already good answer to this question, but I like to demonstrate an important technique. First note that the equation is homogeneous since $$\gcd(k m, k ...
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Proof verification: If $\gcd(n,a) = 1$ and $n\mid ab$ then $n\mid b$.

Another proof of this, but using multisets (you can have repeated elements): Let A, B, N be the multiset that contains the prime factorization of a,b,n. $n \mid ab \implies N \subseteq (A+B)$, but as $...
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0 votes

Prove that either $m$ divides $n$ or $n$ divides $m$ given that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$?

Let $g=gcd(m,n)$, then converting lcm to gcd: $g+\frac{mn}{g}=m+n \iff g^2+mn=g(m+n) \iff g^2-g(m-n)+mn=0 \iff g^2-gm-gn+mn=0 \iff (g-m)(g-n)=0$ This gives that either $g=m$ or $g=n$. WLOG $g=m \...
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1 vote

How to find all natural numbers $n$ such that $a \mid n$ implies $(a+1) \mid (n+1)$?

Let $n$ be a good number. Claim 1. $n$ is not a composite number. Claim 2. $1$ is a good number (which is obvious since $(1+1) \mid 2$). Claim 3. All primes but $2$ are good. In conclusion, we've ...
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2 votes

Collatz conjecture but with $n^2-1$ instead of $3n+1.$ Does the sequence starting with $13$ go to infinity?

About Eric Syder's comment: suppose $ x^2 - 1 = 2^k y$ with $x,y$ odd. We wish to investigate what happens when $y \leq x \; . \; \;$ Note that $\gcd(x+1, x-1) = 2$ because $x \equiv 1,3 \pmod ...
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  • 130k
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Integers starting at 1 and ending at 5 that are divisible by 9.

$a+b+c\equiv 3\bmod 9$, so $abc\equiv 3\bmod 9$ (where $abc$ denotes the 3-digit integer equal to $100a+10b+c$). So $abc$ can take every ninth value between $3$ and $993$ inclusive. You can take it ...
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  • 60k
3 votes
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Finite set of quadratic forms representing all prime numbers

$\newcommand\lgn[2]{(#1\backslash#2)}$ There is no such finite set of forms. Lemma If $f$ is a positive-definite binary quadratic form, and $f$'s discriminant is $D$, and $f$ represents a prime $p$, ...
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  • 2,388
0 votes

Evaluate/asymptotic of the sum $\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$

I get $ \frac{N}{2}(1-2\ln(2))+O(L(N)) $. So we want $m \le N/(2a+1) \lt m+1 $ or $m/N \le 1/(2a+1) \lt (m+1)/N $ or $N/m \ge 2a+1 > N/(m+1) $ or $(\frac{N}{m}-1)/2 \ge a \gt (\frac{N}{m+1}-1)/2 $ ...
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Combinatorial interpretation of Euclid's form for even perfect numbers

@Francis Begbie, how do you come up with that combinatorial form of a perfect number? I'm trying to contact you but that's not possible... Is there a way I can contact you?
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  • 49
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Are $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$?

This is for the question in the title. Let $q$ be a prime dividing both numbers. Then $q\mid p^2+1$, say $p^2+1 = Kq.$ Also note that $q\neq p$, so we must have $$q\mid p^5 -1 = p^5 + p^3 - p^3 - p ...
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  • 29.4k
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A question in proof of Linnik's Theorem in Arithmetic Large Sieve

$\mathcal{P}$ is by definition the set of prime numbers smaller than $N^{1/2}$ such that all positive integers smaller than $N^\epsilon$ are quadratic residues. While $n_p$ is the smallest positive ...
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7 votes

Why does this cycle of 44 show up in the Collatz Conjecture?

There are two types of steps involved in a Collatz sequence: One that multiplies a number by $\frac{1}{2}$, and one that approximately multiplies a number by 3. (The “approximately” part is because ...
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  • 7,376
2 votes

Why does this cycle of 44 show up in the Collatz Conjecture?

Warning: This is more of a comment than an answer and is terribly non-rigorous. First, $44 \approx 2 \pi *7$, so 44 is 'approximate period' of angles in polar coordinates. So we need to show that 44 ...
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  • 2,176
1 vote
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Are $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$?

Assume $\frac{p^2 + 1}{2}$ is not coprime with $\frac{p^{5n}(p^5 - 1)}{2}$, and that $\frac{p^2 + 1}{2} < \frac{p^{5n}(p^5 - 1)}{2}$, so then $\frac{p^2 + 1}{2} \mid \frac{p^{5n}(p^5 - 1)}{2}$. ...
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  • 239
2 votes
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Find solution using infinite descent.

Show that the equation $$x^2 - 3y^2 = 2z^2 \tag1 $$ has no solution in positive integers, by using the argument of infinite descent. First, consider what the argument of infinite descent signifies. ...
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1 vote

Find a polynomial of the form $F(x,y,z)$ of degree $3$ such that $F(a,b,c) = 0 \pmod{5}$ iff $a,b,c= 0 \pmod{5}$

I suppose that this is just a hint. First, the natural environment for this is $\Bbb F_5=\Bbb Z/5\Bbb Z$. You’re looking for a cubic form in three variables with no non-trivial zeros in $\Bbb F_5$. ...
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  • 59.4k
0 votes

Questions in theorem related to primes with fixed modulus

It appears that $\rho(p)$ is the number of solutions to $x^2+1\equiv 0\pmod{p}$. So the statement $\rho(p)=2$ for $p\equiv 1\pmod{4}$ and $\rho(p)=0$ for $p\equiv 3\pmod{4}$ which are relatively ...
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4 votes

Find the invers of $4 \in \mathbb{Z}_5$ (The 5-adic integers)

Hint: $$ \frac{1}{4}= 4+ 3\cdot\frac{5}{1-5}$$
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2 votes
Accepted

What are the fixed points of the arithmetic derivative over the non-negative integers?

(I finally got around to turning my brain on.) Any fixed point must be a prime power. Indeed, if $p \mid n$, then say $n = f(n) = f(pk)$; then $f(n) = p f(k) + k f(p) = p f(k) + k = pk$. Therefore $p ...
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0 votes

What are the fixed points of the arithmetic derivative over the non-negative integers?

I recalled from the video I mentioned in the question that Michael shows that the arithmetic derivative satisfies the power rule: $$f(a^b) = ba^{b-1}$$ Well, if $a = b = p$ where $p$ is any prime, ...
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  • 1,316
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2 questions in the theory of Counting Perfect Squares

For question $2$, notice that $0$ is a perfect square. Now there are $p-1$ numbers remaining, and both $n^2$ and $(-n)^2$ get sent to the same number but are distinct (note $p>2$ for them to be ...
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1 vote
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$M_k(\Gamma_1(N)) = \bigoplus_{\chi \mod N} M_k(N, \chi)$

This requires some basic representation theory (or you can reproduce the representation-theoretic theorems in a special case). Note that $\Gamma_1(N)$ is a normal subgroup of $\Gamma_0(N)$. So take ...
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5 votes
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Is $(1+c^2)^n-\lfloor(1+c^2)^{n/2}\rfloor^2<(1+c^2)^{(n+1)/2}$ true for all integers $c>1$, when $n$ is an odd integer?

I'd argue that it is still possible to do via $\lfloor x\rfloor\geq x-1$: $$\left(1+c^{2}\right)^{n}-\lfloor\left(1+c^{2}\right)^{\frac{n}{2}}\rfloor^{2}\leq\left(1+c^{2}\right)^{n}-\left(\left(1+c^{2}...
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  • 1,071
1 vote

Periodic sequences of integers generated by $a_{n+1}=\operatorname{rad}(a_{n})+\operatorname{rad}(a_{n-1})$

This is basically a comment of how hard the problem is by demonstrating that the sequence starting with $a_1=a_2=p$ with $a_{n+1}=\operatorname{rad}(a_n)+\operatorname{rad}(a_{n-1})$ is equivalent to ...
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  • 1,732
0 votes
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Periodic sequences of integers generated by $a_{n+1}=\operatorname{rad}(a_{n})+\operatorname{rad}(a_{n-1})$

Just a comment, in the intention to recognize patterns giving hints towards analytical arguments about periodicities. In the table below I show the sequences for $N=60$ steps for the primes $p=3..97$ ...
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1 vote
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Notation $(S,*)$ or $(\mathbb S,*)$ for semigroup leading to group $(\mathbb G,*)$

In my experience, the use of blackboard bold often signifies a certain “global uniqueness”, like there is only one set $\mathbb R$ of real numbers. If not that, there will often be at least local ...
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  • 2,651
0 votes

A question in proof of analytic large sieve

Since $|w|^2 = w\overline w$ for any complex number $w$, we have \begin{align*} \biggl| \sum_{ h \pmod p} e(ah /p) Z(p,h) \biggr|^2 &= \biggl( \sum_{ h \pmod p} e(ah /p) Z(p,h) \biggr) \overline{\...
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  • 63.6k
2 votes
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How to rigorously interpret and transform "equal chance" in different ways?

Putting the $100$ balls randomly in the $10$ boxes is equivalent to choosing randomly a mapping from the set of balls to the set of boxes. There are $10^{100}$ such mappings. The event «no box is ...
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  • 12.3k
1 vote

How to rigorously interpret and transform "equal chance" in different ways?

First of all, re Stars and Bars Theory, the two methods that you used do give the same answer. That is $p_{10}(100)$ bijects to determining the number of non-negative integer solutions to $x_1 + \...
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1 vote
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Maximal unramified at p extension of $\mathbb{Q}$

Yes: $\pi_1^{\mathrm{et}}(\mathrm{Spec}(\mathbb{Z}_{(p)})$ is $\mathrm{Gal}(K/\mathbb{Q})$ where $K$ is the maximal extension of $\mathbb{Q}$ unramified at $p$. This is not special to $\mathbb{Z}_{(p)}...
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3 votes
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Field over which CM endomorphism is defined

Yes. Let $E/K:y^2=x^3+ax+b$ such that $End(E)$ is isomorphic to an order $O$ in $\Bbb{Q}(\sqrt{-D})$. For any $\alpha\in O$ let $\phi\in End(E)$ be the "multiplication by $\alpha$" ...
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  • 67.7k
2 votes

Show that there does not exist integers $n_{1}, \ldots, n_{8}$, not necessarily distinct, such that $n_{1}^{4}+\cdots+n_{8}^{4}=1993$.

HINT.-You have $n^4\equiv0\pmod{16}$ when $n$ is even and $n^4\equiv1\pmod{16}$ when $n$ is odd then taking modulo $16$ you have $ n_{1}^{4}+\cdots+n_{8}^{4}\le8$. But $1993\equiv 9\pmod{16}$
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  • 25.3k
4 votes
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Show that there does not exist integers $n_{1}, \ldots, n_{8}$, not necessarily distinct, such that $n_{1}^{4}+\cdots+n_{8}^{4}=1993$.

You have the right idea, but using modulo $16$ instead works better (as also suggested in Arturo Magidin's comment). Similar to modulo $8$, in modulo $16$, every even number raised to the power $4$ is ...
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  • 39.4k
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Units in ring of integers of $5^{th}$ cyclotomic field

Consider $$f: O_K^\times\to \Bbb{Z}[\varphi]^\times, z\to |z|^2$$ where $\varphi= \frac{1+\sqrt5}{2}$. $\ker f = \{ a\in O_K^\times,\forall \sigma\in Hom_\Bbb{Q}(K,\Bbb{C}),|\sigma(a)|=1\} = \mu_{10}$...
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  • 67.7k
1 vote

The set of integers $n$ expressible as $n=x^2+xy+y^2$

Denote $a^2+ab+b^2$ by $L(a, b)$. Numbers of the form $L$ are known as Loeschian numbers. What you are trying to prove is that the set $S$ of Loeschian numbers is multiplicative. Here's a direct proof ...
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  • 2,388
1 vote

Proving that this infinite product is convergent

By partial summation, one can deduce the existence of a constant $C_0$ such that $$ \sum_{p\le z}{w(p)\over p}=K\log\log z+C_0+O\left(1\over\log z\right) $$ As $t\to0^+$, we know $-\log(1-t)=t+O(t^2)$,...
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  • 3,360
3 votes

Prove that $\frac{2022}{n} + 4n$ is a perfect square iff $\frac{2022}{n} - 8n$ is a perfect square

You can greatly reduce the number of cases to check by working modulo $4$. We have $$2022 = 2\cdot 3 \cdot 337 \equiv 2\cdot 3\cdot 1 \pmod 4.$$ Since a square must be $0$ or $1$ mod $4$, we can ...
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  • 5,721
0 votes

Looking for number systems with solutions for $x^2 = 0$ other than $x = 0$

Let $n\geq2$ be an integer, and consider the ring $\mathbb{Z}_{n^2}$ (the integers modulo $n^2$). Then clearly $n<n^2$, so $n$ is not the zero element in $\mathbb{Z}_{n^2}$, but $$n^2\equiv0\pmod{n^...
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  • 2,467
1 vote

Looking for number systems with solutions for $x^2 = 0$ other than $x = 0$

Consider the commutative ring $\Bbb Z/25\Bbb Z$. Then $x^2=0$ has five solutions, namely $$ x=0,5,10,15,20. $$ The ring $\Bbb Z/n\Bbb Z$ is a field if and only if $n$ is prime. The equation $x^2=0$ ...
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