All Questions
34
questions
1
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1
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101
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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 ...
- 83
-1
votes
1
answer
46
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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 ...
- 9
4
votes
0
answers
243
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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 ...
- 234
1
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0
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84
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Find the remainder when the $2006! + \dfrac{4012!}{2006!}$ is divided by $4013$
$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$
Answer: $x=1553.$
Solution: $$2006!+4012!/2006!=x\pmod{4013}$$
$$(2006!)^2 -2006!x+4012!=0\pmod{4013} (*)$$
$$4\cdot (2006!)^2-4\cdot 2006!x+4\cdot 4012!=...
- 11
9
votes
3
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371
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Can $7$ be the smallest prime factor of a repunit?
Repunits are numbers whose digits are all $1$. In general, finding the full prime factorization of a repunit is nontrivial.
Sequence A067063 in the OEIS gives the smallest prime factor of repunits. ...
- 221
0
votes
2
answers
114
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Proving the divisibility of $4[(n-1)!+1]+n$ by $n(n+2)$ in the condition of $n,n+2 \in P$ where $P$ is the set of prime numbers [duplicate]
Let $n$ and ($n+2$) be two prime numbers. If any real value of $n$ satisfies that condition, then prove that $$\frac{4{[(n-1)!+1]}+n}{n(n+2)} = k$$ where $k$ is a positive integer.
SOURCE: BANGLADESH ...
- 1,188
1
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0
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43
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Non-Linear Diophantine Equation in Two Variables [duplicate]
How many solutions are there in $\mathbb{N}\times \mathbb{N}$ to the equation $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}$ ? I could solve till I got to the point where $1995^2$ is equal to the ...
- 344
5
votes
1
answer
224
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An approximation for $1\leq n\leq N$ of the number of solutions of $2^{\pi(n)}\equiv 1\text{ mod }n$, where $\pi(x)$ is the prime-counting function
We denote the prime-counting function with $\pi(x)$ and we consider integer solutions $n\geq 1$ of the congruence $$2^{\pi(n)}\equiv 1\text{ mod }n.\tag{1}$$
Then the sequence of solutions starts as $$...
user243301
0
votes
2
answers
71
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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 $$360 =...
- 689
2
votes
1
answer
142
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The Chinese hypothesis revisited
In the past I tried to get different variations of the so-called Chinese hypothesis, see this Wikipedia (a disproven conjecture).
Today I wanted to combine in an artificious way also Wilson-Lagrange ...
user243301
2
votes
0
answers
505
views
Using factorization to solve modulo arithmetic involving big numbers.
In one of my classes, the following approach was shown to solve modulo operations involving huge numbers:
Problem to solve:
49 10 mod 187.
Approach taken:
Prime factorize $187$. It's factors are ...
- 141
2
votes
1
answer
115
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Prime factors of $5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $
Let $n$ be an integer.
Then any prime factor of
$$ 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 +...
- 16.4k
-1
votes
1
answer
125
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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 $b$
...
- 1,450
7
votes
0
answers
174
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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 $N$...
- 1,450
0
votes
1
answer
124
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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 ...
- 1,450