Questions tagged [prime-factorization]

For questions about factoring elements of rings into primes, or about the specific case of factoring natural numbers into primes.

A natural number is prime if it has no positive divisors besides $1$ and itself. The fundamental theorem of arithmetic states that every natural number $n>1$ can be factored uniquely, up to a reordering of the factors, as a product of distinct prime numbers each raised to some power.

This concept holds in a more general setting though. A ring is called a unique factorization domain (UFD) if every non-unit element can be factored uniquely as a product of prime elements in the ring. The ring of integers is an example of a UFD.

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A simple problem on Finding factors

There's a question wherein we have to find the number of factors of $480$ of the form $8n+4$ where $n \geq 0$. Now, the prime factorisation of $480$ gives $2^5 \times 3 \times 5$. This is to be of the form $8n+4$, i.e. $4(2n+1)= 4x$ (where $x$ is…
Ambica Govind
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How can I prove that the results in the following sequence will always share the same factor?

The following sequence is built of $((A^z)^x -1)/6$ where $z$ increases by $1$ and the conditions for $A$ must be that $(A - 1) \mod 6 = 0$ and for $x$ must be that $x$ is odd and $x > 1$: $((A^1)^x -1)/6 $ , $((A^2)^x-1)/6$ , $((A^3)^x-1)/6$ ,…
Isaac Brenig
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Why is time complexity of factoring expressed for $2^n$ and not $N$?

The question I have is what is in the comment of the question: Why is the time complexity of factorization $2^n$? The original author didn't mention why we are considering the input size in bits and not base 10.
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I have almost no idea how to factor numbers this big.

$14425638854646469646839767613420413647898432138735230192512819$ is the product of two prime numbers. Each factor is an answer. I have to give both factors as answers. How would I get to them and which numbers are them? My knowledge of factors is…
Nicito
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Prime factors of $a$ and $a+1$

Quick question Is there any pattern to how a number's prime factors change when you increase it by $1$? Is there any way you can predict a number's prime factors based on the factors of its previous number?
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What's the point of removing $1$ from the list of primes and having an empty product in the fundamental theorem of arithmetic?

I was reading about how we stopped considering the number $1$ as prime since this would mean the fundamental theorem of arithmetic wouldn't apply any more (since we can repeat the number $1$ over and over again and then the factorization wouldn't be…
Sergio
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Prove that the prime factors $p$ of $n=x^2+y^2$ of the form $p=4m+3$ can only have even exponents.

I'm looking for an elementary proof of the statement in the title. Actually I have already proved that if $-\bar{1}$ is a square then $p\equiv 1 \pmod 4$ or $p=2$. I try to deduce from this statement that if $p=4m+3$ is a prime factor of $n$ then…
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for what $n$, $2^{2^n}-2^{2^{n-1}}+1$ is prime

By checking for first few values, I figured out that the number $$N=2^{2^n}-2^{2^{n-1}}+1$$ is prime for $n=1,2,3$ but for $n=4,5$ it is composite. So are there any specifications on $n$ such that $N$ is prime?
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Help in factorization (Explain if I was wrong)

Task: Factorize $ax^2-a-x^2+x$. And my answer is: $$ a(x^2-1)-x(x-1)=a(x-1)(x+1)-x(x-1)$$ Am I right? If not, explain why and give the correct solution.
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How to prove this sentence?

Suppose $n$ is composite. Consider a prime $q$ that is a factor of $n$ and let $(q^k) || n$. Then $q$ does not divide $$\dbinom{n}{q}$$ Why it is correct?
ido kahana
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An inequality with respect to sum of power of prime factors

Let $L > 1 $ be an integer such that $$ L= p_{1}p_{2}...p_{n} = q_{1}q_2 q_3 ... q_m$$ where $p_i$ are primes which are not necessarily distinct and $q_i$ are arbitrary integers. If I take $L=100$, then $p_1=p_2=2$ and $p_3 = p_4 = 5$ and let $q_1 =…
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Why $r \notin R^*$ if it is a prime?

My professor gave us the following definition: Let $R$ be a commutative ring. $r \in R$ is prime if $r.R$ is a prime ideal. Then he concluded that this definition tells us $r \notin R^*$ but I do not understand why, could anyone explain this to me…
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Prime factors and relation

Consider the relation ~ on $ℕ$ defined as follows: $x$ ~ $y ⇔x * y$ is a square Interpret x ~ y in terms of prime factors of x and y. $ x = {P_1}^{n_1}{P_2}^{n_2}{P_3}^{n_3}{P_4}^{n_4}...{P_a}^{n_a}$ $y =…
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Polynomial in $\mathbb{Z}[x]$ which is irreducible but not prime.

I know that in an integral domain, prime implies irreducible. Moreover, in a principal integral domain, these notions are equivalent. The ring of polynomials $\mathbb{Z}[x]$ is not a principal integral domain (if it were, an irreducible ideal $I$…
inquisitor
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Help proving that $\sum_{d^k|n}\mu(d) =$ 1 if is k-free and 0 otherwise

In class we saw a similar result when $k=2$, and now I'm trying to extend this to an arbitrary $k\in \mathbb{Z}^+$. When I plug in values this identity seems to hold, however I'm unsure how to tackle the proof. This is how I started: Suppose that…
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