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I am quite lucky I am on sickness benefit so I don't have to work. I get quite a lot of enjoyment out of life really.
I don't get it. Why would.................
prove that all possible prime factors will =mn+1..........................?
Neither did I. I think enjoying life is No. 1 priority, otherwise there's no point having more of a bad thing. So maybe, the purpose of life is to enjoy it so that it's better to be alive than not.
I think the meaning of life is to survive and have children. Taking my theory from Darwin- "survival of the fittest". It's good to be alive right? Right. So the only reason you are alive is because your ancestors survived and had children. (If they hadn't done that you wouldn't be here). Therefore for your future offsprings life you must survive and have children. Therefore the purpose of life in general is to survive and have children.
For a mersenne prime............
, all possible prime factors will = where n= prime and m=any multiple..........This will definitely work because of Fermat's little theorem
is always factorable by p, if p is prime and the fact that the pattern of remainders repeats itself so remainder 1 = and n=(p-1)/m, and p=mn+1!!!Thanks so much!
Now I can continue with my mersenne prime theorem.
Hello
I am doing some work on some Mersenne primes and can't figure out the following;
Why is this always divisible by p when p is prime?
A= all the primes below
multiplied togetherI'm not sure but I'm guessing this would be a faster way to compute finding primes. I.e. using Euclid's algorithm.
See............
https://en.wikipedia.org/wiki/Euclidean_algorithm
Take your number, p, which you want to find out if it's prime or not.
Square root it. Multiply all the primes below the square root together.
Minus p from this number and see if there are any common factors between the two. If there is p is not prime, if there isn't p is prime.
Example:
p=51
square root p=7ish
2*3*5*7=210
210-51=159 common factor=3
51 is not prime
p=53
square root p=7ish
2*3*5*7=210
210-53=157 no common factors
53 is prime.
Note: It is impossible if the two numbers add up to a no. factorable by everything, for one to be factorable by a number and the other not. They must either both be factorable or both not.
Don't know what happened there then. It worked for all the ones I tried.
Using a prime number, multiply it by 2.
Separate it into 2 numbers separated by 10x.
Multiply one of them by 10 and add it to the other no.
You should now have another prime but you may need to divide it by 2 until odd.
Example:
13 X 2 = 26
Split into 23 and 3
3 X 10 = 30 plus 23 =53
23 X 10 = 230 plus 3 = 233
Assumptions:
1. There are regular gaps of 4 between primes
2. These gaps of 4 have varying remainders for 4, (either 1 or 3)
A= 3x5x7x11x13x17............(x next prime in series)
x = the next prime above highest prime in A used
p and p+4 are prime
Example:
A= 3x5x7= 105
x=11
A has remainder=1 for 4
2 Primes = 67 and 71
67 and 71 have remainder=3 for 4, which is different from A
(105-67)/2 = 19
(105-71)/2 = 17
Twin primes!
I realise my first post is incomplete so I have made some amendments...............
A= 3, 15, 105, 1155, 15015 x next prime in the series................
p= prime or not factorable by any primes in A
..........And so on until a gap that must be >2 becomes a gap that must be >infinity, there therefore must be an infinite No. of twin primes.
A= 3, 15, 105, 1155, 15015 x next prime in the series................
p= prime or not factorable by any primes in A
..........And so on until a gap that must be >2 becomes a gap that must be >infinity, there therefore must be an infinite No. of twin primes.
Thanks for the post. I have checked out the website. Very interesting!
I have posted this on other topics, but I wanted to make sure everybody has got it!
My theory is simple, Primes have a pattern and I shall tell you what it is.
The pattern is based on primes multiplied together. 2, 6, 30, 210, 2310, 30030.
Then just multiply by the next prime in the series to get, 30030*17=510510 the next number in the series.
From now on you will regard 1 as a prime number as it behaves in much the same way as the primes, but from now on if you see a 1 just remember that it is indeed not a prime.
Okay, let's start with 2.
Write down 2 cross it out, then put a 1 before it. Now just add 2 up to 6........1 3 5. Now times 3 by all no.'s below 2 = 3. Cross it out.
Now let's start with 6. Now add 6 up to 30.............1 5 7 11 13 17 19 23 25 29. Now times 5 by all no.'s below 6 = 5,25. Cross it out.
Now 210. Now add 30 up to 210...........1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83 89 91 97 101 103 107 109 113 119 121 127 131 133 137 139 143 149 151 157 161 163 167 169 173 179 181 187 191 193 197 199 203 209. Now times 7 by all no.'s below 30 = 7,49,77,91,119,133,161,203. Cross them out.
Keep doing this and you will generate all the primes.
Note: In using 210 I have used the primes 2,3,5 and 7. The next prime is 11, so I have only generated the primes up to
p=any number ending in 3
check p is not a square
where a+b is a factor and a-b is another factor
if p is prime you will not find any factors
if p is composite you will find factors.
Example:
p=33
27*33=891
Simpler version
There are no more primes after prime, z.
A=3*5*7*11*13*17*19*23*29*31............................................*z.
p= No. not factorable by any primes in A or 2.
p+/-1=
There are no more twin primes after prime, z.
A=3*5*7*11*13*17*19*23*29*31....................................*z
Ok, cool.
But is that really any better than, say, the Sieve of Eratosthenes?
You would have thought knowing what a and b end in you could determine that a number is composite and therefore not prime, turns out it doesn't seem to work that way.
It wouldn't -- using your notation, x is prime iff (a,b) = (1, x) or (x,1). Even if a or b is 1 modulo 10, that doesn't guarantee that a or b is 1, and for large x, will give you a very large number of possibilities.
I still don't get this post..................? Is 1 modulo 10 a mathematical way of saying, ends in 1.....? I was trying to prove x is composite not prime......? a or b ending in 1 wouldn't guarantee x is prime...?
You were right............it doesn't work.
I made a big mistake:
Primes minus No.'s with remainder e would = (not >1) using my method..........which is not true.
Hi Bob,
Thanks for looking at my posts. I will try to recreate it into simpler style and put it in an algorithm. Might take me a little while. Get back to you later................Primenumbers.
Sorry.
I got totally lost at "but we don't want it to have the same remainders as e".
We have to delete remainders for e otherwise if a had r.=x, then e-a=b: e(with r.x)- a(with r.x)=r.0 for b, so b would not be prime.
Does that help Bob?
Think of it this way;
Way of finding primes=
2m+1
6m+1 or 5
30m+1 or 7 or 11 or 13 or 17 or 19 or 23 or 29
210m + 1 or 11 or 13.....(no.'s not factorable by 2,3,5 or 7)...........209
e might have a remainder if you try to factor 3, 5 or 7.
e will never have a remainder for 2.
1) Let's pretend e does have a remainder for 3. I.e. remainder for 3=x
There will only be 1 no. with r.x in 6.
2) Let's pretend e does have a remainder for 5 also. I.e. remainder for 5=y
There will only be 2 no.'s with r.y in 30.
3) Let's pretend e does have a remainder for 7 also. I.e. remainder for 7=z
There will only be 8 no.'s with r.z in 210.
You can now see that minusing these no.'s from no. of primes there will still be enough left to make a and b.