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TIL...... [modified]

Scheduled Pinned Locked Moved Algorithms
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  • N NeverHeardOfMe

    ...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

    L Offline
    L Offline
    Lost User
    wrote on last edited by
    #4

    Try

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

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

    N R 2 Replies Last reply
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    • L Lost User

      Try

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

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

      N Offline
      N Offline
      NeverHeardOfMe
      wrote on last edited by
      #5

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

      L L 2 Replies Last reply
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      • N NeverHeardOfMe

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

        L Offline
        L Offline
        Lost User
        wrote on last edited by
        #6

        :confused:

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

        N 1 Reply Last reply
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        • L Lost User

          :confused:

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

          N Offline
          N Offline
          NeverHeardOfMe
          wrote on last edited by
          #7

          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)?

          L 1 Reply Last reply
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          • N NeverHeardOfMe

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

            L Offline
            L Offline
            Luc Pattyn
            wrote on last edited by
            #8

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

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

            Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

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

            N 1 Reply Last reply
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            • N NeverHeardOfMe

              ...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

              L Offline
              L Offline
              Luc Pattyn
              wrote on last edited by
              #9

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

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

              Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

              N 1 Reply Last reply
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              • L Luc Pattyn

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

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

                Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

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

                N Offline
                N Offline
                NeverHeardOfMe
                wrote on last edited by
                #10

                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!

                L 1 Reply Last reply
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                • L Luc Pattyn

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

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

                  Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

                  N Offline
                  N Offline
                  NeverHeardOfMe
                  wrote on last edited by
                  #11

                  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

                  1 Reply Last reply
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                  • N 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)?

                    L Offline
                    L Offline
                    Lost User
                    wrote on last edited by
                    #12

                    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

                    N 1 Reply Last reply
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                    • N 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!

                      L Offline
                      L Offline
                      Luc Pattyn
                      wrote on last edited by
                      #13

                      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

                      Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

                      N 1 Reply Last reply
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                      • L 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

                        N Offline
                        N Offline
                        NeverHeardOfMe
                        wrote on last edited by
                        #14

                        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 ...

                        1 Reply Last reply
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                        • L 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

                          Please use <PRE> tags for code snippets, they preserve indentation, and improve readability.

                          N Offline
                          N Offline
                          NeverHeardOfMe
                          wrote on last edited by
                          #15

                          Now you're just confusing me....

                          1 Reply Last reply
                          0
                          • N NeverHeardOfMe

                            ...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

                            _ Offline
                            _ Offline
                            _Erik_
                            wrote on last edited by
                            #16

                            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.

                            N M 2 Replies Last reply
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                            • _ _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.

                              N Offline
                              N Offline
                              NeverHeardOfMe
                              wrote on last edited by
                              #17

                              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.

                              _ 1 Reply Last reply
                              0
                              • N 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.

                                _ Offline
                                _ Offline
                                _Erik_
                                wrote on last edited by
                                #18

                                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

                                N 1 Reply Last reply
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                                • _ _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

                                  N Offline
                                  N Offline
                                  NeverHeardOfMe
                                  wrote on last edited by
                                  #19

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

                                  1 Reply Last reply
                                  0
                                  • _ _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.

                                    M Offline
                                    M Offline
                                    MarkLoboo
                                    wrote on last edited by
                                    #20

                                    3 cheers !!:thumbsup:

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

                                    1 Reply Last reply
                                    0
                                    • L Lost User

                                      Try

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

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

                                      R Offline
                                      R Offline
                                      redbones
                                      wrote on last edited by
                                      #21

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

                                      1 Reply Last reply
                                      0
                                      • N NeverHeardOfMe

                                        ...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

                                        D Offline
                                        D Offline
                                        dbaechtel
                                        wrote on last edited by
                                        #22

                                        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

                                        1 Reply Last reply
                                        0
                                        • N NeverHeardOfMe

                                          ...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

                                          D Offline
                                          D Offline
                                          Daniel Pfeffer
                                          wrote on last edited by
                                          #23

                                          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

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