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**Primenumbers**- Replies: 0

p=Any number

A=2x3x5x7x11x13x17........ up to

m=Any multiple

A-pm= A number factorable by a factor of p

if p has one.To get A-pm, go through the primes, subtracting p as many times as you like.

Example:

p=129

A=2x3x5x7x11

2x3x5x7=210 210-129=81

81x11=891 891-(129x6)=117

A-129m=117

117 will have a factor the same as p if it has any.

117/129=39/43 129 is NOT prime (Common denominator =3)

In this way we can find out if p is composite without ever having to use a number

. Might be useful for computers.**Primenumbers**- Replies: 0

Basically I take my number p, square root it, Rd. Up to the nearest prime, then the next prime greater than that -1= largest prime gap below p.

Example:

p=130

Rd. Up to nearest prime= 13

Next prime after that = 17

17-1=16

Largest prime gap <130 = 16 (Correct)

This works because the greatest number of composites between two primes occurs when factors are not combined. So what could have been two composites is actually just one, like 15=3x5. To create the greatest possible number of composites I start at 2 not 0. 0 has an infinite number of prime factors, and so the greatest gap between the next repeat will occur after 0. Starting with the smallest composite which is NOT combined factors, I move up. Deleting all numbers factorable by primes less than the square root. The first time I attain TWO primes is when I reach the second prime after the square root. So this -1 is the gap required to create two non-composites with greatest possible occurrence of composites.

Largest prime gap <p =

Rd. Up to second nearest prime -1.**Primenumbers**- Replies: 2

A new discovery about prime numbers is that they are not truly random. Prime numbers that are not factorable by 2 or 5 only end in 1,3,7or 9. If prime numbers were truly random we would expect a prime number ending in 1 to be followed by another prime ending in 1 about 25% of the time but apparently using the first billion prime numbers the likelihood is more like 18%. This is a bias and they discovered biases for prime numbers ending in 3,7 and 9 as well.

Am I missing something here? Don't we know that there is a bias?

We know that any prime not factorable by 2, 3 or 5 = 30m + 1 or 7 or 11 or 13 or 17 or 19 or 23 or 29. Surely we can work out biases from this, for example a number ending in a number is never followed by another number ending in the same number. I.e. A number ending in 1 never follows another number ending in 1. This means the other numbers are more likely. And that's what Soundararajan and Oliver discovered; an 'anti-sameness' bias.

zetafunc wrote:

It isn't particularly difficult to understand -- I can explain it to you if you just tell me what it is that you need help with.

I get the beginning but not the end of it where you've put (-1/p)=-1.

zetafunc wrote:

Just to make sure, this is what you want to know, correct?

-You are considering only numbers of the form any non-zero positive integer.

-You would like to know something about the prime factorisations of these numbers.

First point is correct but I just wanted to show that by using

we can reduce the number of possible factors which should help compute primes faster. And now you have shown that will never be factorable by the primes of the form so thank you. I thought about 50% of primes could not possibly be factors but now you're saying that not only would there be 50%, but that there would be a bias to more than 50% due to Chebyshev's bias. Have I got this right?I am also still trying to work out if there's anything else we can use this scenario for to do with primes. If there was a wheel of prime factors that could factor

, it would be 20m + 1 or 9 or 13 or 17, not including 2 and 5.zetafunc wrote:

You can immediately rule out any primes such that whenever is not itself prime.

But after you have ruled out primes where

all that is left are p's For instance n=12 =145 rule out/divide by 5 and you get 29, a primezetafunc wrote:

You can also rule out primes

by Euler's criterion (or more specifically, by the corollary I mentioned in my previous post).

I don't get Euler's criterion, and probably never will it's just too advanced for me, don't worry. So what you're basically saying is we can rule out possible prime factors where p=4m+3 for

?zetafunc wrote:

may only be factorable by factors of where x<nPrimenumbers wrote:Potential factors for

may only be factored by previous values ofForgive my ignorance, but I have no idea what this means. Could you explain this?

zetafunc wrote:

Because.............

Let p= potential factor

Let p=n-y

(n-y)(n+y) =

Therefore remainder when is divided by p =And remainder when

is divided by p =That's correct. But I still don't know what a potential factor is (or what you need this part for).

y=n-x

zetafunc wrote:

y will be less than p because y=n-p and p>0

You mean y will be less than n, not p.

Yes

zetafunc wrote:

p will not be greater than n because we are only concerned with primes

So I am guessing by "potential factor" you mean "prime factor"? Also, assuming you want n^2 + 1 to be composite (which isn't true in general, as your next post shows), you need a ≤ sign.

Basically x<n because we are only concerned with factors below the square root, and because y=n-x, y cannot=n because x>0.

zetafunc wrote:

This should reduce potential factors for

You mean reduce the possibilities for the prime factors of n^2 + 1?

Yes

zetafunc wrote:

I have tried up to n=43. Prime factors <44 that were missing were 3, 7, 11, 19, 23, 31 and 43 which exactly halved the no. of potential prime factors. Any value for n to infinity will not be factorable by these no.'s.

then they will never occur because ones above p are just repeats of ones <p.

The factors repeat themselves every +p value for n. So if they don't occur beforeI still don't really understand what you are trying to do, from your post or from your examples in the post above. If you're wondering why 3, 7, 11, 19, 23, 31 and 43 don't appear in the prime factorisations of n^2 + 1, then this is because of quadratic reciprocity. (Let me know if you haven't come across this term before, and I'll be happy to explain it.) In other words, consider the congruence for p = 3 mod 4:

but this doesn't have any solutions, because if , then , where is the quadratic reciprocity symbol. (Which is just another way of saying that -1 isn't a square modulo p, where p is 3 modulo 4 -- in other words, the congruence has no solutions for p = 3 mod 4.)

Basically I am trying to make it so that instead of a computer having to try to factor all the primes below the square root for an integer, it only has to try and factor a lot less. Also because the factors repeat themselves this could make things easier. Maybe could do a wheel or something? No, I have never come across Euler' Criterion and I don't get it.

**Primenumbers**- Replies: 1

I was going to go to a Taekwondo class but then I found out online that martial arts are completely useless in a real fight apparently. And I wanted self-defence training.

As I thought about this I realised it was completely true..................

I am trained in Judo, (no punching and kicking), and my identical twin is one belt below black in karate. We had lots of fights when I was mentally ill and it would usually end in a standoff because we were both pretty evenly matched. I didn't use any of my throws or holds and that means my twin's brown belt in karate was useless against me who would actually be a white belt in karate. How amazing is that!

Apparently things like boxing, wrestling and MMA are better because you actually get hit and experience the real ruff-and-tumble of a real fight.

7 is prime so only factorable by 1 or 7.

the only squares less than 6 are 1 and 4 therefore a=1, r=2

a=7, r=0

73 is prime so only factorable by 1 or 73. 73 too big so must be 1. Only smaller than 73 is r= 1 or 2. 1 is too small so a=1, r=2.

Examples: [Formulas followed by prime factorization, as you can see factors 3,7,11,19,23,31 and 43 are missing]

n=1

n=2 5

n=3 10=2,5

n=4 17

n=5 26=2,13

n=6 37

n=7 50=2,5

n=8 65=5,13

n=9 82=2,41

n=10 101

n=11 122=2,61

n=12 145=5,29

n=13 170=2,5,17

n=14 197

n=15 226=2,113

n=16 257

n=17 290=2,5,29

n=18 325=5,13

n=19 362=2,181

n=20 401

n=21 442=2,13,17

n=22 485=5,97

n=23 530=2,5,53

n=24 577

n=25 626=2,313

n=26 677

n=27 730=2,5,73

n=28 785=5,157

n=29 842=2,421

n=30 901=17,53

n=31 962=2,13,37

n=32 1025=5,41

n=33 1090=2,5,109

n=34 1157=13,89

n=35 1226=2,613

n=36 1297

n=37 1370=2,5,137

n=38 1445=5,17

n=39 1522=2,761

n=40 1601

n=41 1682=2,29

n=42 1765=5,353

n=43 1850=2,5,37

**Primenumbers**- Replies: 9

New Theorem:

Potential factors for

Because.............

Let p= potential factor

Let p=n-y

(n-y)(n+y) =

Therefore remainder when is divided by p =

And remainder when is divided by p =

y will be less than p because y=n-p and p>0

p will not be greater than n because we are only concerned with primes

This should reduce potential factors for

and reduce computing space for finding primes.I have tried up to n=43. Prime factors <44 that were missing were 3, 7, 11, 19, 23, 31 and 43 which exactly halved the no. of potential prime factors. Any value for n to infinity will not be factorable by these no.'s.

The factors repeat themselves every +p value for n. So if they don't occur before

**Primenumbers**- Replies: 3

Fermat's Last Theorem states that no three positive integers a, b, and c satisfy the equation

for any integer value of n strictly greater than two.Let

where n >2Then

must equal:must have opposite remainders for

Adding y to one and minusing y from another will keep them opposite if they are already opposite. So let's see if the two squares have opposite remainders for .

If we add them together they should=0 if they are opposite.

No!

Fermat's Last theorem is True.

I don't know, it's just a Maths degree, starting 2017. Doing Foundation Year and Placement Year.

Maths now.

I got into Uni to do maths! I am going to do that now.

Q: Why did the angry Jedi cross the road?

A: To get to the Dark Side.

No way! That's not awesome.

Thirty and counting!

Can be written as....

And we know

So,

569,673=(1x2)+(2x4)+(4x6)+(6x8)+(10x12)+(12x14)+(16x18).

Well usually you'd do

where x=(2-1)(3-1)(5-1)(7-1) and it would be factorable by 2,3,5,7........But if you use a composite Ie. it won't be factorable by 3 and 7.

In this way you can find out if the base is prime or not in one simple calculation!

Oh! Fair enough.