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1 vote
1 answer
101 views

distribution of square roots of unity $mod n$ | Factoring with inverse pair

I am writing a proof related to the RSA cryptosystem, specifically showing that given an inverse pair $d, c$ under multiplication mod $\phi(N)$, where $$ dc \equiv 1 \pmod{\phi(N)}, $$ there exists a ...
FieldHouser's user avatar
4 votes
0 answers
99 views

Minimum $k$ for which every positive integer of the interval $(kn, (k+1)n)$ is divisible by at least one prime number less than $n$

As a continuation of this question relating the Minimum $k$ for which every positive integer of the interval $(kn, (k+1)n)$ is composite and this other one on the divisibility of numbers in intervals ...
Juan Moreno's user avatar
  • 1,190
3 votes
1 answer
58 views

Divisibility of numbers in intervals of the form $[kn,(k+1)n]$ [duplicate]

I have checked that the following conjecture seems to be true: There exists no interval of the form $[kn, (k+1)n]$ where each of the integers of the interval is divisible by at least one of the ...
Juan Moreno's user avatar
  • 1,190
0 votes
0 answers
42 views

About the proof of reduction of factoring to order finding

Inside the book I'm following there's a theorem, used to prove the factoring algorithm, which states: Suppose $N = p_{1}^{\alpha_1}p_{2}^{\alpha_2}\dots p_{m}^{\alpha_m}$ is the prime factorization of ...
Francesco Greco's user avatar
-1 votes
1 answer
46 views

Finding common modulo

given these two modulo equations $c_1 = m_1^a (\mod n)$, $c_2 = m_2^a (\mod n)$ Where '$a$' is prime and $n$ is a product of two primes, and the only unknown is $n$, is it possible to solve for $n$? I ...
Kyle's user avatar
  • 9
2 votes
1 answer
68 views

How to solve $x^x \equiv 0 \pmod y$

Given a constant y, I am trying to find the smallest value for x that satisfies the equation $x^x = 0 \mod y$. So far I have been able to determine that $x$ is equal to the product of all the prime ...
cw123's user avatar
  • 21
3 votes
1 answer
147 views

Is the number of ways to express a number as sum of two coprime squares same as number of solution of $x^2+1\equiv0\pmod n$

The number of representations of $n$ by sum of 2 squares is known as sum of square function $r_2 (n)$. It is known that if prime factorization of $n$ is given as $$2^{a_0}p_1^{a_1}p_2^{a_2}\cdots q_1^{...
didgogns's user avatar
  • 3,589
4 votes
0 answers
243 views

Proper divisors of $P(x)$ congruent to 1 modulo $x$

Let $P(x) $ be a polynomial of degree $n\ge 4$ with integer coefficients and constant term equal to $1$. I am interested in Polynomials $P(x) $ such that for a fixed positive integer $b$, there are ...
ASP's user avatar
  • 234
1 vote
1 answer
50 views

Probability a random number $M$ is not a factor of $N$

Let $N$ be some positive integer and let $S := \lbrace 1, 2, \cdots, \log^2(N) \rbrace$ (pretending at $\log^2(N)$ is an integer). Suppose $M$ is randomly chosen from the set $S$. The goal is to use ...
spektr's user avatar
  • 549
1 vote
0 answers
84 views

Find the remainder when the $2006! + \dfrac{4012!}{2006!}$ is divided by $4013$

$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$ Answer: $x=1553.$ Solution: $$2006!+4012!/2006!=x\pmod{4013}$$ $$(2006!)^2 -2006!x+4012!=0\pmod{4013} (*)$$ $$4\cdot (2006!)^2-4\cdot 2006!x+4\cdot 4012!=...
user825769's user avatar
0 votes
0 answers
55 views

Why is this not sufficient proof of the divisibility of $\binom{p}{j}$ by $p$.

In my text book there's an example of a proof on why $\binom{p}{j}$ is divisible by $p$, with $p$ prime, for $0<j<p$. Firstly, it shows that $$\binom{p}{j}=p\frac{(p-1)!}{j!(p-j)!}$$ From this ...
lafinur's user avatar
  • 3,468
-1 votes
1 answer
161 views

Computing (quickly) the multiplicity of a (prime) divisor

Question I have a fixed, prime d and an n < 2⁶⁴, and I want not only to compute whether d ...
nicoo's user avatar
  • 99
1 vote
2 answers
162 views

Finding all no-congruent primitive roots $\pmod{29}$

Finding all no-congruent primitive roots $\pmod{29}$. I have found that $2$ is a primitve root $\pmod{29}$ Then I found that is it 12 no-congruent roots, since $\varphi(\varphi(29)) = 12$ Then I ...
magnus's user avatar
  • 113
0 votes
1 answer
44 views

Does there always exist a prime $q\equiv3\mod 4$ that divides $p+a^2$ with $p\equiv1$ mod 4

Let $p$ be a prime such that $p\equiv1\mod4$. Is it true that there will always exist a prime $q$ that satisfies $q\vert(p+a^2)$ and $q\equiv3\mod4$, for some integer $a$? I have tried proceeding by ...
user520830's user avatar
0 votes
1 answer
351 views

Find integral solution using congruence modulo.

Find integral solution to $a^3 - 1100 =b^3$ using modular arithmetic. No integral solutions for this exist, so how to prove using modular arithmetic? Earlier I had asked about $a$ and $b$ being ...
Shawn Kelly's user avatar
1 vote
1 answer
128 views

For $n \ge 4$ find a factorization $n^2 - 3n + 1 = ab$ where $a \lt n$ and $b \lt n$.

Update: We can use Willie Wong's argument to justify the definition of a 'truth cutoff' function, $\quad \psi: \{3,4,5,6, \dots \} \to \{4,5,6,7, \dots \}$ For convenience we start with a ...
CopyPasteIt's user avatar
  • 11.5k
1 vote
2 answers
221 views

Find the least odd prime factor of $155^8+1$

Find the least odd prime factor of $155^8+1$. How do I do this without using Wolfram Alpha or something?
Silverleaf1's user avatar
1 vote
0 answers
122 views

Sum of members from multiplicative group of prime order $k$ modulo prime $P$? $c$ in: $\sum_{n=1}^{k} (g^n \bmod P) = c \cdot P$ ($g$ prime order $k$)

Let $P$ be a prime ($>2$) and $g$ a value between $2$ and $P-2$. Let $M$ be the set of numbers which can be generated with $g$: $$M = \{g^n\bmod P, \text{ with } 0 < n <P \}$$ If $g$ is a ...
J. Doe's user avatar
  • 77
2 votes
1 answer
106 views

Integer Factor Congruence

Given an integer $N$, with unknown prime factors $f_1$, $f_2$ ... $f_n$, and given unique integers $k_1$, $k_2$ ... $k_n$, with $\sqrt{N} \geq k_i>2$ for all $i$ such that $$f_1 \equiv 1\pmod {k_1}$...
Ruan Sunkel's user avatar
9 votes
3 answers
371 views

Can $7$ be the smallest prime factor of a repunit?

Repunits are numbers whose digits are all $1$. In general, finding the full prime factorization of a repunit is nontrivial. Sequence A067063 in the OEIS gives the smallest prime factor of repunits. ...
Vincent Luo's user avatar
3 votes
1 answer
126 views

The proof of $(n+1)!(n+2)!$ divides $(2n+2)!$ for any positive integer $n$

Does $(n+1)!(n+2)!$ divide $(2n+2)!$ for any positive integer $n$? I tried to prove this when I was trying to prove the fact that ${P_n}^4$ divides $P_{2n}$ where $n$ is a positive integer, where $P_{...
hteica's user avatar
  • 438
0 votes
1 answer
196 views

Calculate all prime numbers $x$, where $x^{18} - 1$ is divisible by $28728$

Question: Calculate all prime numbers $x$, where $x^{18} - 1$ is divisible by $28728$ Apparently, the answer is all prime numbers except $2, 3, 7,$ and $19.$ I did some prime factorisation and found ...
Alexander B's user avatar
1 vote
1 answer
80 views

Given 2 functions $f(x)$ and $g(x)$ find a value where $g(x)$ divides $f(x)$ meaning $f(x) = 0 \mod g(x)$

Problem: Given 2 functions $f(x) = 2^{p-1} + x*p$ and $g(x) = 2 * x * p + 1$ find the values where $f(x) = 0 \mod{g(x)}$, where $p$ is a prime number and $x$ is a non negative integer in the range $1,...
Erick84mm's user avatar
0 votes
1 answer
154 views

How to find the prime factors when knowing some congruence?

In order to factorize the integer $N = 67591$, choose a factor base $\{2,3,5\}$ and four congruences: $24256^2 \equiv 2^9 \cdot 3^4(mod\ N)$; $59791^2 \equiv 2^2 \cdot 3^4\cdot 5^2(mod\ N)$; $23541^2 \...
Ganlin Zh's user avatar
0 votes
2 answers
114 views

Proving the divisibility of $4[(n-1)!+1]+n$ by $n(n+2)$ in the condition of $n,n+2 \in P$ where $P$ is the set of prime numbers [duplicate]

Let $n$ and ($n+2$) be two prime numbers. If any real value of $n$ satisfies that condition, then prove that $$\frac{4{[(n-1)!+1]}+n}{n(n+2)} = k$$ where $k$ is a positive integer. SOURCE: BANGLADESH ...
Anirban Niloy's user avatar
1 vote
2 answers
79 views

Show that any prime divisor of $x^4+x^3+x^2+x+1$, with $x\in\mathbb{N}$, is $5$ or $1$ mod $5$

We can write the "polynomial" as follows: $$x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.$$ For even $x=2y$, we have that $x^5-1=(2y)^5-1=32y^5-1\equiv1$ mod $5$. For odd $x=2y+1$, we have that $(2y+1)^5-1\...
Algebear's user avatar
  • 1,674
1 vote
0 answers
43 views

Non-Linear Diophantine Equation in Two Variables [duplicate]

How many solutions are there in $\mathbb{N}\times \mathbb{N}$ to the equation $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}$ ? I could solve till I got to the point where $1995^2$ is equal to the ...
Anonymous's user avatar
  • 344
1 vote
1 answer
89 views

Find X in the Equation

I'm not a mathematician and I have forgotten about some basics in mathematics. I have this equation: $$x^y \pmod z = w$$ Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $...
alyssaeliyah's user avatar
0 votes
0 answers
79 views

Finding the smallest prime factor of $\sum_{a=1}^N a^{k}$

This question is linked to my previous question, but I wanted a clearer explanation. Suppose we have a huge number of that type with a huge $k$. $$\sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k},$$ ...
alienflow's user avatar
  • 349
0 votes
1 answer
165 views

Find the smallest positive prime divisor of ...

Problem: That's a problem I have found on the web. I didn't understand the solution: Why?? Given solution: How all this sequence has been transformed into $$33-{\lfloor {33\over p}\...
alienflow's user avatar
  • 349
7 votes
2 answers
2k views

What is the multiplicative order of a product of two integers $\mod n$?

Standard texts prove that $\textrm{ord}_n(ab)=\textrm{ord}_n(a)\,\textrm{ord}_n(b)$ when $\textrm{gcd}(\textrm{ord}_n(a),\textrm{ord}_n(b))=1$. What if they are not relatively prime? Here $\textrm{...
Conifold's user avatar
  • 11.8k
1 vote
0 answers
39 views

Baby steps to factorization of a large composite number via modular arithmetic

Given a huge composite number with unknown prime factors and a much smaller prime modulus P, is there a way (short of the Herculean task of factorization) to identify the residue classes mod(P) of the ...
Bert Barrois's user avatar
3 votes
1 answer
113 views

What does it mean for $\sigma^N$≡$\sigma^e$ (mod N) for arbitrary a?

Full disclosure: I am trying to solve a homework problem. As part of a homework problem, we were given a large semiprime $N$, which was used as both the modulus and the public key and asked to ...
Richard Hum's user avatar
5 votes
1 answer
224 views

An approximation for $1\leq n\leq N$ of the number of solutions of $2^{\pi(n)}\equiv 1\text{ mod }n$, where $\pi(x)$ is the prime-counting function

We denote the prime-counting function with $\pi(x)$ and we consider integer solutions $n\geq 1$ of the congruence $$2^{\pi(n)}\equiv 1\text{ mod }n.\tag{1}$$ Then the sequence of solutions starts as $$...
user avatar
0 votes
2 answers
71 views

Find all $x \in \mathbb Z_{360}$ such that $x^2 ≡ 0 \pmod{360}$

Find all $x \in \mathbb Z_{360}$ such that $$x^2 ≡ 0 \pmod{360}.$$ I know that this means to find all $x$ such that the result divides into $360$ evenly. I also know the prime factorization of $$360 =...
C.Math's user avatar
  • 689
2 votes
1 answer
142 views

The Chinese hypothesis revisited

In the past I tried to get different variations of the so-called Chinese hypothesis, see this Wikipedia (a disproven conjecture). Today I wanted to combine in an artificious way also Wilson-Lagrange ...
user avatar
2 votes
0 answers
505 views

Using factorization to solve modulo arithmetic involving big numbers.

In one of my classes, the following approach was shown to solve modulo operations involving huge numbers: Problem to solve: 49 10 mod 187. Approach taken: Prime factorize $187$. It's factors are ...
Vivek Maran's user avatar
0 votes
1 answer
80 views

Can at most $3$ distinct primes divide $n^3-n$, for infinitely many $n$?

My assignment asks me to prove that there are only finitely many $ n\in\mathbb{N} $ such that the prime factorization of $n^3-n$ is of the form $p_1^{r_1} p_2^{r_2} p_3^{r_3}$ for $p_i$ primes and $...
user529653's user avatar
3 votes
0 answers
114 views

If 13 does not divide m, then prove that $m^4+8$ is not a cube of an integer [closed]

My question is how can we prove that $m^4 + 8$ not a cube of an integer if $m$ can not be divided by 13. What I have done so far: By Fermat’s Little Theorem: \begin{align} m^{p-1} &\equiv 1 \...
Singh Chief's user avatar
2 votes
1 answer
115 views

Prime factors of $5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $

Let $n$ be an integer. Then any prime factor of $$ 5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $$ Must be congruent to 1 mod 10. Also Let $n$ be an integer. Then any prime factor of $$ 5 n^4 - 10 n^3 +...
mick's user avatar
  • 16.4k
-1 votes
1 answer
125 views

Found $a^2\equiv b^2(\mod RSA\_1024)$ What are the chances?

Due to the size of the numbers, I am writing them as a code. Below are $a$ and $b$ ...
Ilya Gazman's user avatar
  • 1,450
7 votes
0 answers
174 views

I found a way to calculate Quadratic min mod $N$, but why does it work?

I am trying to factor $N$ using Dixon's factorization method, so I am looking at the equation: $$a^2\equiv b(\mod{N})$$ If I am able to find $b$ that is a perfect square, I will be able to factor $N$...
Ilya Gazman's user avatar
  • 1,450
0 votes
1 answer
124 views

Find integer $x$ such that $x^2 \mod {1799832043}$ is divisible by $67610791$

Find integer $x$ such that $x^2 \mod {n}$ is divisible by $p$ For values $n = 1799832043, p = 67610791$ I have been using Tonelli-Shanks algorithm to solve this and it works for small primes with ...
Ilya Gazman's user avatar
  • 1,450
0 votes
1 answer
28 views

How hard is it to find $k = j(ed-1)$ in RSA signature

In RSA signatures the below formulas come into play: $N=rq$ where $r$ and $q$ are large primes $ed=1 \mod{\varphi(N)}$ where $e$ is small number with no common division with $\varphi(N)$ $a=m^d \...
Ilya Gazman's user avatar
  • 1,450
0 votes
2 answers
86 views

Why $\gcd(a,a+N) =1$

By experiment, I notice that $$\gcd(a,a+N) =1$$ Where $N$ is a big composite integer number that is hard to factor and does not have a common divisor with $a$. And $a$ is a positive big integer that ...
Ilya Gazman's user avatar
  • 1,450
0 votes
0 answers
42 views

How to simplify $2^{a\left(2^{bc}-1\right)}-1\mod{N}$

How to simplify $$2^{a\left(2^{bc}-1\right)}-1\mod{N}$$ Where a,b,c are big integers with relatively small primes and N is a big integer that is hard to factor(it does not have small primes). I have ...
Ilya Gazman's user avatar
  • 1,450
1 vote
0 answers
299 views

Pohlig-Hellman algorithm

I'm trying to use the Pohlig-Hellman algorithm to solve for $x$ where $15^x=131(\bmod 337)$. This is what I have so far: prime factors of p-1: $336=2^4*3*7$ $q=2: x=2^0*x_0+2^1*x_1+2^2*x_2+2^3*x_3$ ...
Matt Robbins's user avatar
1 vote
6 answers
2k views

Sum of positive divisors if and only if perfect square

let n be a positive odd integer, prove that the sum of the positive divisors of n is odd if and only if n is a perfect square. I know that based on the prime factorization theory that every integer n ...
Skrrrrrtttt's user avatar
1 vote
0 answers
198 views

Proving a number is a factor of a in (a mod N)

Suppose you had a very large composite number, which is the product of a couple large primes. Let's call this $a = c_1c_2c_3c_4$ where $c_1,c_2,c_3$ and $c_4$ are all primes. If you calculated $d = ...
Joe Thomas's user avatar
0 votes
0 answers
67 views

Is $(2^a - 3^b)$ mutually prime with $(2^a - ((2^a)^{-1} \mod 3^b ) - ((3^b)^{-1} \mod 2^a))$?

Is $(2^a - 3^b)$ mutually prime with $(2^a - {{[2^a]}^{-1}}\mod 3^b + [3^b]^{-1}\mod 2^a)$, provided that $a\geq b$ and $b>0$? By $[2^a]^{-1}\mod 3^b$ I mean "the multiplicative inverse of $2^a\...
Joe Slater's user avatar

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