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Questions tagged [prime-numbers]

Prime numbers are natural numbers greater than 1 not divisible by any smaller number other than 1. This tag is intended for questions about, related to, or involving prime numbers.

A prime number (or a prime) is an element of the greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number ... The fundamental theorem of arithmetic establishes the central role of primes in :

Any integer greater than 1 can be expressed as a product of primes that is unique up to ordering.

Here you get the first 50 millions of primes.

The concept of prime numbers is extended in ring theory, where an element $p$ of a ring $R$ is prime if and only if whenever $p\mid ab$, then $p\mid a$ or $p\mid b$.

One can easily see that this extends the definition of prime numbers in the natural numbers.

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Shapes and sizes of finite sets of prime numbers

Knowing that $p$ is prime enables us to rule out the possibility that $p+2$ and $p+4$ are both prime, except in the one trivial case that $p=3$, since at least one of $p,\ p+2,\ p+4$ is divisible by $3$. But in some cases, $p,\ p+2,\ p+6$ are all…
Michael Hardy
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Solving $2x \equiv 1 \pmod{p}$ where $p$ is an odd prime

Solve $2x \equiv 1 \pmod{p}$ where $p$ is an odd prime. I'm really stuck on this one.
KarlX
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Using Bertrand's Postulate

Using Bertrand's postulate which states: For every integer $n \geq 1$ there is a prime number p such that $n
sarahusher
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Find all ordered triples $(x,y,z)$ of prime numbers satisfying equation $x(x+y)=z+120$

This question was from my Math Challenge II Number Theory packet, and I don't get how to do it. I know you can distribute to get $x^2+xy=z+120$, and $x^2+xy-z=120$, but that's as far as I got. Can someone explain step by step?
Jason Chen
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Prime number proof

Given that $n>2$. Prove that if $2^n-1$ is prime then $2^n+1$ is composite or vice versa. I looked on wikipedia on Fermat number and Mersenne prime, but I still don't know how they work.
user148361
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Is a Mersenne-Prime always of the form $3n + 1$?

Examples are: $M_3 = 7 = 3\times 2 + 1$ $M_5 = 31 = 3\times 10 + 1$ $M_7 = 127 = 3\times 42 + 1$ $M_{13} = 8191 = 3\times2730 + 1$
Jerry
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Is there a asymptotic formula for product of primes?

$$P(x)=\prod_{p\leq x}p$$ As you can see P(x) represents the product of primes which are not greater than x. Is there a asymptotic formula for this?
esege
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Why there are no other known Fermat primes.

Fermat primes are prime numbers of the form $2^{2^n} + 1$: $$3,~5,~17,~257,~65537$$ There are no other known Fermat primes. But why?
user142289
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Is the product of two primes ALWAYS a semiprime?

I know by definition, a semi-prime has factors that are prime numbers. But what I'm unsure of, is if there is ever a case where the product of two prime numbers results in number with factors OTHER than the original two prime numbers? Or will any…
Eddie
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Disjoint subset/prime question

If $p$ is a prime, which integers $n$ is it possible to split the set consisting of $\{1,2,...,n\}$ into $p$ disjoint (meaning they have no element in common) subsets where the sum of the integers in each subset has the exact same value?
user137151
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Check if any number of this form is composite

I got this problem to solve saying : Given that $n>2$ and $n$ is a natural number, prove that $y = 2^n + (-1)^{n+1}$ is a composite number. I took the above and I did the following: $$y = 2^n + (-1)^n(-1)^1 = 2^n - (-1)^n$$ and because of the…
Takis
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For $n>2$ is no primes followed by even powers?

I was looking at the question Is $7$ the only prime followed by a cube? and I was running through some of the low powers of $n^m-1$ I tried $m=2$ and got that $n^2-1=(n+1)(n-1)$ I reasoned that $n+1>1$ because otherwise $n-1<0$ and cannot be prime.…
kleineg
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Generalization of the prime number theorem to other fields

Is there a similar statement to the prime number theorem in other rings like $\mathbb{Z}[i]$ or $\mathbb{Z}[\omega]$.
Mayank Pandey
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how often do we find $p^m - q^n= \pm2$ for primes $p,q$ and $m,n > 1$

if one of the integers $m,n$ is $1$ it does not seem too difficult to find examples of odd primes satisfying: $$|p^m-q^n| = 2$$ so suppose $\min(m,n)>1$, and call (just for the purpose of this question) primes satisfying the condition mentioned…
David Holden
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Instantly Factor a Semiprime of Any Size?

recently I've been looking at the interesting problem of RSA encryption and attempting to understand what it's so hard to find the factors then I came across this page. Can anyone comment on it's validity? http://www.naturalnumbers.org/Qfactor3.html
Derek James
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