TIL...... [modified]

[Radhakrishnan G]

what is that "n" it is 1 for p = 5??

[NeverHeardOfMe]

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 :)

[Lost User]

Try

p ^ 2 = 1 + 24 * (p - 6)

Just say 'NO' to evaluated arguments for diadic functions! Ash

[NeverHeardOfMe]

OK... p^2 - 24*p + 143 = 0 So..?

[Lost User]

:confused:

Just say 'NO' to evaluated arguments for diadic functions! Ash

[NeverHeardOfMe]

Me :confused: 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)?

[Luc Pattyn]

a quadratic equation like that has at most two solutions, so it is hardly a way to discover lots of primes. :)

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modified on Thursday, November 4, 2010 12:05 PM

[Luc Pattyn]

that is correct, even when n needs to be 1/8 for p=2. :-D

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[NeverHeardOfMe]

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!

[NeverHeardOfMe]

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! :-O

[Lost User]

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

[Luc Pattyn]

you'll have more luck with

x2 + x + 41

for natural values of x. :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles] Nil Volentibus Arduum

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[NeverHeardOfMe]

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 ...

[NeverHeardOfMe]

Now you're just confusing me....

[_Erik_]

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.

[NeverHeardOfMe]

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.

[_Erik_]

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

[NeverHeardOfMe]

Yes, sorry - I was forgetting that p is prime! Time to knock off for the day I think...

[MarkLoboo]

3 cheers !!:thumbsup:

All are born right-handed. Only gifted few overcome it. There's NO excuse for not commenting your code.

[redbones]

actually that's not correct; when p=7 then by your formula we would have 49=1+24*1 = 25