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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 ...
10 votes
1 answer
1k views

When is $(p-2)!-1$ power of $p$ if $p$ is prime?

If $p$ is prime, for what values of $p$ is $(p-2)!-1$ a power of $p$? I know how to solve that when $p<5$ then $(p-1)!+1$ can be written as power of $p$.
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 ...
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 ...
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\...
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 ...
16 votes
6 answers
16k views

prime divisor of $3n+2$ proof

I have to prove that any number of the form $3n+2$ has a prime factor of the form $3m+2$. Ive started the proof I tried saying by the division algorithm the prime factor is either the form 3m,3m+1,3m+...
2 votes
2 answers
1k views

Calculating $n$ mod $m$ given the prime factorization of $n$

Say I have the prime factorization of a large integer $n$. $$n=p_1^{a_{1}}\cdot p_2^{a_{2}}\ldots p_k^{a_{k}}$$ However, I do not have $n$ itself. How do I calculate $n$ mod $m$, given only $n$'s ...
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 ...
1 vote
0 answers
175 views

Check whether a no has exactly two Prime Factors [duplicate]

Is there a way to check whether an integer has exactly two prime factors(can be same) that is , it can be expressed as a product of two prime nos ? EDIT: I dont want to find the factors i just want a ...
-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 ...
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 ...
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^{...
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 ...
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 ...

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