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1 vote
1 answer
99 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
  • 1,180
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
  • 1,180
1 vote
0 answers
95 views

The chunking aspect of repunit prime factors [closed]

While others have already mentioned the divisibility of decimal repunits, $$m := \frac{{10}^n - 1}{9}.$$ ...
RARE Kpop Manifesto's user avatar
0 votes
0 answers
41 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
  • 3,599
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
  • 244
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
  • 3,408
-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
161 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

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