Prime Definition and 777 Threads

Hello! I'm asked to factor 314000 by hand. The answer key says that is is ##2^4\cdot 5^3\cdot 157##, but I honestly have no idea how to factor it by hand. I know that I can check all prime numbers up to ##\sqrt{314000}## but that would not be doable by hand obviously. I did try to use the...

I would like to know if my answer is correct and if no ,could you correct.But it should be right I hope:

How should I write an account of prime numbers in arithmetic progressions? Assuming this account should be in the form of an essay of at least ## 500 ## words. Should I apply the formula ## a_{n}=3+4n ## for ## 0\leq n\leq 2 ##? Can anyone please provide any idea(s)?

Caution I'm not a mathematician. In short, long time ago I calculated prime number gaps just for fun expecting an almost uniform distribution of the frequency of the gaps 2, 4, 6, ... . Instead the frequency showed a series of maxima and minima and I was confused. Later Professor emeritus Oskar...

Proof: Let ## p ## be the prime divisor of two successive integers ## n^{2}+3 ## and ## (n+1)^{2}+3 ##. Then ## p\mid [(n+1)^{2}+3-(n^{2}+3)]\implies p\mid (2n+1) ##. Observe that ## p\mid (n^{2}+3) ## and ## p\mid (2n+1) ##. Now we see that ## p\mid [(n^{2}+3)-3(2n+1)]\implies p\mid...

im thinking i should just integrate (binominal distribution 1-2000 * prime probability function) and divide by integral of bin. distr. 1-2000. note that I am looking for a novel proof, not just some brute force calculation. (this isn't homework, I am just curious.)

Hi PF! I created a function ##R(x)## that gives the gap between the largest two primes less than or equal to ##x##. To define it, I used this property: $$\pi(x+R(x))=\pi(x)+1$$ Which is true since the ##x## distance between ##\pi(x)## and ##\pi(x)+1## is ##R(x)##. If we solve for ##R(x)## we...

Proof: By Dirichlet's theorem, we have that if ## a ## and ## d ## are two positive coprime numbers, then there are infinitely many primes of the form ## a+nd ## for some ## n\in\mathbb{N} ##. Let ## n\geq 1 ## be a natural number. Now we consider the arithmetic progression ## 10^{n+1}k+1 ##...

I wrote a program that implements the pattern and finds the primes automatically. It worked up to 70 million then it crashed because program holds data in RAM so it can be fixed. It found all the primes up to 70 million and found no exception. I won't explain the pattern because its so...

From my research I have found that since Fermat proved his last theorem for the n=4 case, one only needs to prove his theorem for the case where n=odd prime where c^n = a^n + b^n. But I am not clear on some points relating to this. For example, what if we have the term (c^x)^p, where c is an...

Proof: Let ## p ## be the prime divisor of two successive integers ## n^2+3 ## and ## (n+1)^2+3 ##. Then ## p\mid [(n+1)^2+3-(n^2+3)]\implies p\mid 2n+1 ##. Since ## p\mid n^2+3 ## and ## p\mid 2n+1 ##, it follows that ## p\mid a(n^2+3)+b(2n+1) ## for some ## a,b\in\mathbb{Z} ##. Suppose ## a=4...

Proof: Suppose ## p ## and ## p^{2}+8 ## are both prime numbers. Since ## p^{2}+8 ## is prime, it follows that ## p ## is odd, so ## p\neq 2 ##. Let ## p>3 ##. Then ## p^{2}\equiv 1 \mod 3 ##, so ## p^{2}+8\equiv 0 \mod 3 ##. Note that ## p^{2}+8 ## can only be prime for ## p=3 ##. Thus ##...

Proof: Let ## n>3 ## be an integer. Applying the Division Algorithm produces: ## n=3q+r ## for ## 0\leq r< 3 ##, where there exist unique integers ## q ## and ## r ##. Suppose ## n ## is prime. Then ## n=3q+1 ## or ## n=3q+2 ##, because ## n\neq 3q ##. Now we consider two cases. Case #1...

Proof: Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##, which is an odd integer that is not greater than ## n ##. Let ## n>1 ## be an integer. Since ## n! ## is even, it follows that ## n!+1 ## is odd. Thus ## 2\nmid (n!+1) ##. This means every prime factor...

Proof: Assume to the contrary that ## n!-1 ## is not prime. That is, ## n!-1 ## is composite. Let ## p ## be a prime factor of ## n!-1 ## such that ## n<p ## for some ## n\in\mathbb{N} ## where ## n>2 ##. Since ## p\mid (n!-1) ##, it follows that ## p\nmid n! ##. This means ## p\nleq n ##...

Proof: Suppose for the sake of contradiction that any composite three-digit number must have a prime factor not less than or equal to ## 31 ##. Let ## n ## be any composite three-digit number such that ## n=ab ## for some ## a,b\in\mathbb{Z} ## where ## a,b>1 ##. Note that the smallest prime...

if I select two integers at random between 1 and 1,000, what is the probability that their sum will be prime?

Proof: Suppose for the sake of contradiction that ## \sqrt{p} ## is not irrational for any prime ## p ##, that is, ## \sqrt{p} ## is rational. Then we have ## \sqrt{p}=\frac{a}{b} ## for some ## a,b\in\mathbb{Z} ## such that ## gcd(a, b)=1 ## where ## b\neq 0 ##. Thus ## p=\frac{a^2}{b^2} ##...

Proof: Consider all primes ## p\leq\sqrt{701}\leq 27 ##. Note that ## 701=2(350)+1 ## ## =3(233)+2 ## ## =5(140)+1 ## ## =7(100)+1 ## ## =11(63)+8 ## ## =13(53)+12 ##...

I just saw that one of the ways of calculating Pi uses the set of prime numbers. This must sound crazy even to people who understand it, is it possible that this can be explained in terms that I, a mere mortal can understand or it is out of reach for non mathematicians?

## 1234=2\cdot 617 ## ## 10140=2\cdot 2\cdot 3\cdot 5\cdot 13\cdot 13 ## ## 36000=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5\cdot ## Are the answers above correct? Or do I need to put them in canonical form as below? ## 1234=2\cdot 617 ## ## 10140=2^{2}\cdot 3\cdot 5\cdot...

Proof: Suppose ## p\neq5 ## is an odd prime. Applying the Division Algorithm produces: ## p=10q+r ## where ## 0\leq r\leq 10 ##, where there exist unique integers ## q ## and ## r ##. Note that ## p\equiv 1, 3, 7 ## or ## 9 ## mod ## 10 ##. This means ## p^{2}\equiv 1, -1, -1 ## or ## 1 ## mod...

Proof: Note that all primes less than 50 will divide 50!, because each prime is a term of 50!. Applying the Fundamental Theorem of Arithmetic produces: Each term k of 50! that is non-prime has a unique prime factorization. Since 48, 49 and 50 are not primes, it follows that all primes...

Proof: Suppose that p is a prime and ##p \mid a^n ##. Note that a prime number is a number that has only two factors, 1 and the number itself. Then we have (p*1)##\mid##a*## a^{(n-1)} ##. Thus p##\mid##a, which implies that pk=a for some k##\in\mathbb{Z}##. Now we have ## a^n ##=## (pk)^n ##...

Proof: Suppose p##\geq##5 is a prime number. Applying the Division Algorithm produces: p=6k, p=6k+1, p=6k+2, p=6k+3, p=6k+4 or p=6k+5 for some k##\in\mathbb{Z}##. Since p##\geq##5 is a prime number, it follows that p cannot...

Proof: Suppose that p is a prime and 3p+1=n^2 for some n##\in\mathbb{Z}##. Then we have 3p+1=n^2 3p=n^2-1 3p=(n+1)(n-1). Since n+1>3 for ##\forall...

Proof: Suppose p is a prime such that p=n^2-4. Then we have p=n^2-4=(n+2)(n-2). Note that prime number is a number that has only two factors, 1 and the number itself. Since n+2>1 for ##\forall n \in...

Proof: Suppose p is a prime such that p=n^3-1. Then we have p=n^3-1=(n-1)(n^2+n+1). Note that prime number is a number that has only two factors, 1 and the number itself. Since n^2+n+1>1 for ##\forall...

Build the least common multiple of A, B, and C Then write the prime factorization of the least common multiple of A, B, and C. $A = 2 \cdot 3^2 \cdot 7 \cdot 13 \cdot 23^8$ $B = 2 3^5 \cdot 5^9 \cdot 13$ $C = 2 \cdot 5 \cdot 11^8 \cdot 13^3$ $\boxed{?}$ ok this only has a single answer...

Build the least common multiple of A and B a. write the prime factorization of the least common multiple of A and B. $A=2\cdot 3^2\cdot 5\cdot 7^3\cdot 11^3\cdot 13^2$ $B= 3^2 \cdot 5 \cdot 7^2 \cdot 11^2$ $\dfrac{2\cdot \cancel{3^2}\cdot \cancel{5\cdot 7^2} 7\cdot \cancel{11^2} 11\cdot...

I don't know where. to even begin inserting numbers here. I know the Temperature would be -5 for the first T and 36 for T sub zero but I do not know how to solve this. I never had Calculus.

let p∈Z a prime how can I show that p is a prime element of Z[√3] if and only if the polynomial x^2−3 is irreducible in Fp[x]? ideas or everything is well accepted :)

I hope this is the right place to ask this. I was clicking the Sun a few days ago with my beginner scope and a DSLR. The scope is a 60mm aperture f/12 refractor and the DSLR is a Canon SL3 (APS-C 6000x4000). First I used a 20mm eyepiece (35x) to view the sun, saw it in all its glory, the...

If you go to "The Abel Prize Interview 2016 with Andrew Wiles" on YouTube, there is a statement made by Andrew Wiles beginning at about 4:10 and ending about 4:54 where he mentions there are some new number systems possible where the fundamental theorem of arithmetic does not hold. I don't...

i do not seem to understand part ##ii## of this problem...mathematical induction proofs is one area in maths that has always boggled me :oldlaugh: let ##n=3, p=7, ⇒m=4## therefore, ##7=(4-3)(4+3)## ##7=1⋅7## ##1, 7## are integers...##p## is prime. i am attempting part ##iii## in a moment...

$\tiny{apc.1.1.13}$ The number 1,001 is the product of the primes 7, 11, and 13 Knowing this, What is the prime factorization of 30,030? a, ${3 \cdot 7\cdot 10\cdot 13}$ b. ${30\cdot 7\cdot 11\cdot 13}$ c. ${2 \cdot 5\cdot 7\cdot 11\cdot 13}$ d. ${3\cdot 7\cdot 10\cdot 11\cdot 13}$ e...

https://www.biorxiv.org/content/10.1101/2021.01.08.425835v1.full Can prime editing fix every mutation in every type of cell in the body?. From what I read in the article the editing efficiency of prime editing using adeno-associated virus is 1.82%, so what prevent us from repeating the same...

From @fresh_42's Insight https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

Dear PF Forum, Can someone help me with the algorithm for finding a very large prime number? In RSA Encryption (1024 bit? 2048?, I forget, should look it up at wiki for that), Private Keys is a - two prime number packet. Now, what I wonder is, what algorithm that the computer use to find those...

Find the minimum value for $x$ where $x\in \mathbb{N} $ and $x^2-x+11$ can be written as a product of 4 primes, which are not necessarily distinct.

Hi there, bored physicist here. I wondered about factorization so I made some plots and a model fit to satiate my curiosity, but it only made me more curious. Now I just want the answers! I didn't know which prefix to use. "Hobbyist" seemed more appropriate but we don't have that so I went with...

I get 2^5 x 3^80 Am I correct?

How can I prove that Zp is a vector space if and only if p is prime

Here is my attempt When we raise both sides to the power (p-1)/2, we get x^(p-1)= -1^[(p-1)/2](modp) Looking at p=3(mod4), the possible values of p are {3, 7, 11, 19, 23, 31...}. Putting these values of p into (p-1)/2 we get odd integers. {1, 3, 5, 9, 11, 15...}. So we have x^(p-1) =...

Summary:: Interested in the history of the proof. Consecutive integer numbers are always relatively prime to each other. Does anyone know when this was proved? Was this known since Euclid's time or was this proved in modern times?

$\tiny{412.00.1.12}$ Show that $5n+3$ and $7n+4$ are relatively prime for all n. $$ax + by = 1$$ $\begin{array}{ll} \textit{let} &a=5n+3 \textit{ and } b=7n+4\\ \textit{then} &(5n+3)x + (7n+4)y = 1\\ \textit{compute}&(7n+4)=(5n+3)+(2n+1)\\ &(5n+3)=2\cdot(2n+1)+(n+1)\\...

Let's say the function is of a form F(x), where F(n) gives us the value of the n-th prime number, for all values of n. And it computes and discovers subsequent prime numbers directly without using any form of brute-force computation.If such a function is discovered, how ground-breaking would it...

Hi, I want to know if the algorithm to find prime number has running time O(sqrt(N)) or O(log^2N). log^2N is better than sqrt(N). Also for large values can it become a pseudo polynomial algorithm? Zulfi.

I know that the prime factorization theorem predicts that a prime number raised to an integer power will never be equal to another prime number raised to a different power. But does this apply to real number powers? For example, suppose there is a prime number raised to some real value, could...

Hello all, We know that following formulas failed to produced all prime numbers for any given whole number ##n##: ##f(n) = n^2 - n - 41##, failed for ##n = 41~(f = 1681)## ##g(n) = 2^(2^n) + 1##, failed for ##n = 5~(g = 4,294,967,297)## ##m(n) = 2^n - 1##, failed for ##n = 67~(m =...