|
...an extraordinary fact - probably known to half of you, but nevertheless I thought I'd share it. It's not an alogorithm, but never mind!
The square of any prime number >= 5 can be expressed as 1 + a multiple of 24
or
If p is prime, then p ^ 2 = 1 + 24 * n, for some n
Don't know about you, but I think that's quite amazing. Why is it true at all, and given that it is, where does 24 come from?! (Pity it's not 42 really!)
[edit] When I ask "where does 24 come from?" I am not asking for a mathermatical proof - I can look that up myself. It's more just a metaphysical musing...)
modified on Thursday, November 4, 2010 6:37 AM
|
|
|
|
|
what is that "n" it is 1 for p = 5??
|
|
|
|
|
Yes, for
p=5, n=1
p=7, n=2
p=11, n=5
p=13, n=7
If you can work out a forumla that spits out each value of n that produces the "next" prime, you'll be able to retire early
|
|
|
|
|
Try
p ^ 2 = 1 + 24 * (p - 6)
Just say 'NO' to evaluated arguments for diadic functions! Ash
|
|
|
|
|
OK...
p^2 - 24*p + 143 = 0
So..?
|
|
|
|
|
Just say 'NO' to evaluated arguments for diadic functions! Ash
|
|
|
|
|
Me too - ome of us is being a bit dumb here (most likely me...)
I just re-wrote the equation you posted... what does it prove? Is there a typo in it (yours)?
|
|
|
|
|
No typo as far as I am aware. I just thought you wanted to know what was the value of 'n' in your original question.
Just say 'NO' to evaluated arguments for diadic functions! Ash
|
|
|
|
|
Ah, got you - except that it doesn't work for all values of p
p=5, n=1 != p-6
p=7, n=2 != p-6
p=11, n=5 ok
p=13, n=7 ok
p=17, n=12 != p-6
...
|
|
|
|
|
a quadratic equation like that has at most two solutions, so it is hardly a way to discover lots of primes.
|
|
|
|
|
We have a saying round here: "Don't tell I, tell 'e"... especially as that particular one doesn't even have any (real) solutions at all!
|
|
|
|
|
you'll have more luck with
x<sup>2</sup> + x + 41
for natural values of x.
|
|
|
|
|
Now you're just confusing me....
|
|
|
|
|
actually that's not correct; when p=7 then by your formula we would have 49=1+24*1 = 25
|
|
|
|
|
that is correct, even when n needs to be 1/8 for p=2.
|
|
|
|
|
Actually, that's QI.. it also "works" for p=3 (n=1/3) - so in fact, the *only* value of p for which n is neither an integer nor a proper fraction is 4.[edit] OOPS what an idiot! since when was 4 a prime!
|
|
|
|
|
Well, I know you are not asking for a mathematical proof, but like I am much better at maths than metaphisics...
Where does 24 come from?
Since p is prime and greater than 5, p cannot be an even number. So:
p2-1=(p-1)*(p+1).
Like p is odd, both (p-1) and (p+1) are even, so we can say:
p2-1=2a*2b=4ab.
Like (p-1) and (p+1) are two consecutive even numbers, one of them must be multiple of 4, so 2a is multiple of 4 or 2b is multiple of 4, so we can express this like:
p2-1=4a*2c=8ac
or
p2-1=4b*2d=8bd
I will examine just one of these two cases becouse they are symmetrical. Now, like (p-1) and (p+1) are two consecutive even numbers, one of them must be multiple of 3, so:
p2-1=8a*3d=24ad
or
p2-1=8c*3e=24ce
And there it is.
|
|
|
|
|
You haven't explained the last step correctly, and the statement
_Erik_ wrote: Now, like (p-1) and (p+1) are two consecutive even numbers, one of them must be multiple of 3
is wrong.
|
|
|
|
|
It is not wrong, becouse p is a prime number, what means:
if p mod 3 = 1 then (p-1) mod 3=0, so (p-1) is multiple of 3.
if p mod 3 = 2 then (p+1) mod 3=0, so (p+1) is multiple of 3.
if p mod 3 = 0 then p would not be prime and we would not be following our premise
|
|
|
|
|
Yes, sorry - I was forgetting that p is prime!
Time to knock off for the day I think...
|
|
|
|
|
3 cheers !!
All are born right-handed. Only gifted few overcome it.
There's NO excuse for not commenting your code.
|
|
|
|
|
WolramAlpha say that the equation p^2 = 24*d*e+1
is a hyperboloid of one sheet
which means that it is of the form x^2/a^2 + y^2/b^2 - z^2/c^2 = 1
|
|
|
|
|
For p = 5, p^2 = 24 * 1 + 1
All primes > 5 have the form p = 30*k + l, where l is 1, 7, 11, 13, 17, 19, 23 or 29. All other values of l have a common factor > 1 with 30, and therefore cannot produce primes.
p^2 = 900*k^2 + 60*k*l + l^2
Rewrite the first 2 terms as 60*( 15*k^2 + k*l ). The sum in the parentheses is always even, so the entire expression is divisible by 24.
We are left with l^2. By inspection:
1^2 = 24 * 0 + 1
7^2 = 24 * 2 + 1
11^2 = 24 * 5 + 1
13^2 = 24 * 7 + 1
17^2 = 24 * 12 + 1
19^2 = 24 * 15 + 1
23^2 = 24 * 22 + 1
29^2 = 24 * 35 + 1
QED
|
|
|
|
|
Daniel Pfeffer wrote: All primes > 5 have the form p = 30*k + l, where l is 1, 7, 11, 13, 17, 19, 23 or 29. All other values of l have a common factor > 1 with 30, and therefore cannot produce primes.
That's quite an assertion... not saying it's wrong, but it can't be s starting point for a proof of anything; it has to be proved first. Surely.
|
|
|
|
|
NeverHeardOfMe wrote: Daniel Pfeffer wrote:
All primes > 5 have the form p = 30*k + l, where l is 1, 7, 11, 13, 17, 19, 23 or 29. All other values of l have a common factor > 1 with 30, and therefore cannot produce primes.
That's quite an assertion... not saying it's wrong, but it can't be s starting point for a proof of anything; it has to be proved first. Surely.
I have simply performed the first three steps of the Sieve of Erathostenes, eliminating all numbers divisible by 2, 3 or 5, as follows:
30 is the product of the first three primes (2 * 3 * 5).
All integers may be represented as n = 30*k + l where k is any integer, and l is any integer in the range { 0..29 }.
Of these integers:
Any integers of the form 30*k + {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28} are divisible by 2.
Any integers of the form 30*k + {0, 3, 6, 9, 12, 15, 18, 21, 24, 27} are divisible by 3.
Any integers of the form 30*k + {0, 5, 10, 15, 20, 25} are divisible by 5.
When all integers known to be divisible by 2, 3, or 5 are removed, we are left with 30*k + {1, 7, 11, 13, 17, 19, 23, 29} that MAY be prime.
|
|
|
|
|