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    THERE IS NO SECURE TEST FOR PRIMALITY

    Ing. Pier Francesco Roggero, Dott. Michele Nardelli, Francesco Di Noto

    Abstract

    In this paper we examine in detail a completely new test for the primality of

    prime number and how all these tests behave in the mystery of prime

    numbers.

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    Index:

    1. PRIMALITY TEST ............................................................................................................................. 31.1 SUM OF THE DIGITS OF NUMBER N .............................................................................. 41.2 PRODUCT OF THE DIGITS OF THE NUMBER N ........................................................... 51.3 ANALYSIS OF PAIRS S e P ................................................................................................ 71.4 FURTHER INVESTIGATIONS ........................................................................................... 91.5 CONNECTION WITH PERFECT NUMBERS ..................................................................14

    2. CONCLUSIONS................................................................................................................................163. ANALYSIS OF 1229 PRIMES LESS THAN 10000........................................................................184. REFERENCES ..................................................................................................................................67

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    1. PRIMALITY TEST

    Aprimality test is an algorithm that allows to decide whether a given number is prime

    or not.

    We try to build one.

    Given a number N with the last digit ends in 1, 3, 7 or 9 we add up its digits in modulo

    9 (as is done with the proof of 9) and then multiply each digits in modulo 9.

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    1.1 SUM OF THE DIGITS OF NUMBER N

    As the number N can also be written in polynomial form

    N = an10n+ + a110

    1+ a010

    0= ai10

    i

    We have

    airepresents the sum of all its digits.

    Now the sum of its digits modulo 9 represents the REMAINDER of the division by 9,and then we can write:

    S ai mod 9

    From the sum are automatically removed 9s and the digits that give as the sum 9 or

    multiple, in that its congruence is 0, resulting in a neutral element in the sum that may

    well be omitted.

    If the sum S is equal to 0, 3 or 6 the number is divisible by 3.

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    1.2PRODUCT OF THE DIGITS OF THE NUMBER N

    Similarly, here we have:

    N = an10n+ + a110

    1+ a010

    0= ai10

    i

    The product of all its digits is given by:

    airepresents the product of all its digits.

    Now the product of its digits in module 9 represents, also in this case, the remainder ofthe division by 9, and then we can write:

    P ai mod 9

    In the product if it appears in the digits a 0 or a 9 or digits that give the product as 9

    (3*3, 3*3*3, ) or multiple (6 * 3 6 * 6, ...), it is useless to continue to multiply hisfigures because P = 0

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    From the product P can derive the following interesting table:

    TAB 1:

    P N

    Last digit of

    the prime

    number

    0At least 1 digit of N is 0 or 9 or 2 digits or more of

    N are a multiple of 9(3*3, 6*3, 3*3*6)1, 3, 7, 9

    3, 6 Only 1 digit of N is 3 or 6 1, 3, 7

    1, 4, 7 At least 1 digit of N is 1, 4, 7 1, 7

    2, 5, 8 At least 1 digit of N is 2, 5, 8 1, 7

    Note:

    You cannot have a prime number N with last digit equal to 9 if its P = 3 or 6, because ifthe last digit is 9 P values 0.

    If P = 1, 2, 4, 5, 7 or 8, the last digit of the first number N is 1 or 7, because in this case

    in the digits of N never appear either 0, 3, 6 or 9.

    If P = 3 or 6 in the prime number N appears ONLY ONE digit that is 3 or 6.

    All odd numbers N of the form 6k + 1, potentially prime numbers, or multiples orpowers of primes, which contain a 0, a 9, two 3's, a 6 and a 3, not necessarily

    consecutive, have P = 0, since multiple 9, because 3 * 3 = 9, 6 * 3 = 18 multiple of 9,

    while a 0 cancels the product of the digits of N.

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    1.3ANALYSIS OF PAIRS S e P

    Analyzing the S and P pairs of 1229 prime numbers less than 10,000, some pairs are

    not present.

    More specifically, the following 8 pairs S and P are NOT present:

    S P1 1

    5 1

    7 1

    7 2

    1 4

    2 4

    4 4

    4 7

    In total there are 9 x 9 = 81 pairs, where 27 have S = S = 0, 3 and 6, which are multiples

    of 3. There remain then 54 but ONLY 8 are not eligible to give prime numbers less than10,000.

    Thus, at least for N odd numbers lower than 10 000, such pairs are good only for a

    deterministic test of "non primality".

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    But are we sure that continuing in our analysis also these 8 pairs do not emerge withprime numbers greater than 10,000?

    In fact here are some examples of prime numbers greater than 10,000 that "cover" the

    remaining pairs of S and P.

    TAB. 2

    Primenumber S P

    11251 1 1

    11287 1 4

    111827 2 4

    12451 4 4

    11281 4 7

    11111 5 1

    12877 7 1111121 7 2

    For pairs S and P (2,4) e (7,2) should even go to the prime numbers greater than

    100,000

    These pairs are veryrare but are present.

    The test is probabilistic but, like all other similar tests, inefficient to have certainty.

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    1.4FURTHER INVESTIGATIONS

    Prior to the following table for the first 1229 prime numbers up to 10 000, ie, (10 000)

    = 1229, we make some example with primes of the form 6k + 1

    Table 6k +1 to 25, except for the initial number 1, to understand the general trend of S

    and P, to extending it to larger numbers in case of any possible regularity

    In red numbers multiple of 3, which then will be eliminated, since they are never primenumbers, except for the initial 3.

    In blue the numbers of the form 6k + 1 which include prime numbers, and their products

    and powers, that our test should not obviously give as prime, in black all other even

    numbers and not prime, except for the initial 2 in the first column.

    6k 4S P

    6k-3S P

    6k-2S P 6k-1S P

    6kS P 6k+1S P

    2

    2 2

    3

    3 3

    4

    4 455 5

    6

    6 677 7

    8

    8 8

    9

    9 9

    10

    1 0112 1

    12

    3 2134 3

    14

    5 4

    15

    6 5

    16

    7 6178 7

    18

    9 8191 9

    202 0 213 2 224 4 235 6 246 8 25 compos.7 1

    26

    8 3

    27

    9 0

    28

    1 7292 0

    30

    3 0314 3

    32

    5 6

    33

    6 0

    34

    7 3

    35 compos.

    8 6

    36

    9 0371 3

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    Now we have the following sums S and products P, for each of the six numeric forms,

    which contain all the numbers except the first initial of the form 6 * 0 +1 = 0 +1:

    6k -2

    S P

    6k-3

    S P

    6k-4

    S P6k-5S P

    6k

    S P6k+1S P

    2 2 3 3 4 4 5 5 6 6 7 7

    8 8 9 9 1 0 2 1 3 2 4 3

    5 4 6 5 7 6 8 7 9 8 1 9

    2 0 3 2 4 4 5 6 6 8 7 18 3 9 0 1 7 2 0 3 0 4 3

    5 6 6 0 7 3 8 6 9 0 1 3

    For our purposes only affect the columns in blue in which we note that the digits S are

    repeated with the sequence 5 2 8 in column 6k -1 e 7 4 1 in column 6k + 1,while for the products P such repetition does not exist or is not known, since a table

    should be much longer, what we might see later.

    The cyclicity of S is due to the fact that if a number N is added to a multiple k 'of 9,with k = k*1,5 where k is the k of6k + 1 returns to the original numerical root, for the

    prime numbers, we are interested in this work, we have, starting from 5, 5 + 18 = 23,

    with S = 2 +3 = 5 original, etc.

    3+6 0 2*9 =18 = three lines, which is why every three lines repeats the same value of S.

    For the products P it is more difficult.For primes omirp, and those swappable, they all have the same pair, S and P, being the

    sum and produced two commutative operations.

    Eg. omirp numbers 37 and 73, with the same root number

    S = 3 + 7 =7 + 3 = 10 = 1 + 0 =1 eP = 3*7= 7*3 =21 = 2+1=3

    We also note, and it is logical that in the column 6k + 1 S increased by 2 compared to

    the S in the column 6k -1, being essendo 6k + 1 (6k -1) = 6k +1 - 6k +1 =1+ 1 = 2,

    and taking into account the form if S +2 is two digits, eg. 8 in the third and sixth row, 8

    +2 = 10 and 1 + 0 = 1, in fact 8 + 2 (mod 9) = 1.

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    Table of prime numbers up to 100 form 6 k -1 and relative values of S and P

    Prime number of form6k -1

    S cyclic: 8 5 2, 8 5 2 P not cyclic

    5 5 5

    11 2 1

    17 8 7

    23 5 629 2 9

    35 composite, last digit 5 8 6

    41 5 4

    47 2 1

    53 8 6

    59 5 9

    65 composite, last digit 5 2 3

    71 8 7

    77 composite = 7*11 5 483 2 6

    89 8 9

    95 composite, last digit 5 5 9

    97 2 9

    ..

    We note, however, that some pairs of S and P, for example 8 and 6 are both prime

    numbers (53), which for composite numbers, (35), as well as 5 e 9 for the first prime

    number 59 that for the composite 95 .Therefore, under this aspect, the prime numbers are not privileged by particular pairs of

    S and P.

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    The same goes for prime numbers of the form 6k + 1

    Prime number of form6k +1

    S cyclic: 7 4 1, 7 4 1 P not cyclic

    7 7 7

    13 4 3

    19 1 9

    25 composite, last digit 5 7 131 4 3

    37 1 3

    43 7 3

    49 composite = 7*7 4 9

    55 composite, last digit 5 1 7

    61 7 6

    67 4 6

    73 1 3

    79 7 985 composite, last digit 5 4 4

    91 composite =7*13 1 9

    97 7 6

    Also here we have obviously the same phenomenon, but with the pair 1 9, for both theprime number 19, for both the composite number 91.

    But the following table, the result of an algorithm, shows all the 1229 prime numbers up

    to 10000, with its pairs S and P, where we can observe the pair 8 and 6 occurs for someof them.

    Here, however, are obviously not present the composite numbers with pair 8 and 6, as

    we have seen for the composite number 35 (8, 6) for the prime number 53 (permutation

    of 35), 467, 647 (permutation of 467), 2861, 5651, while the pair (8, 0) is present for a

    large number of prime numbers.

    Then each pair would have its frequency, which ranges from a minimum of attendance

    (zero prime numbers for the two pairs (2, 4), (7. 2) and high, as for example the couple

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    above (8, 0), there are already 16 appearances up to 1000 as the pair (8, 6) is presentonly 5 times up to 10 000.

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    1.5 CONNECTION WITH PERFECT NUMBERS

    The mathematical Sicilian Filippo Giordano noted that all perfect numbers, except for

    the initial 6, have as the sum S = 1, for example, to 28, S = 2+8 = 10 = 1+ 0 =1, per

    496, S = 4+9+6 = 19 = 1+ 9 = 10 = 1,etc.. (Ref. 2, "The odd perfect numbers", already

    on our website. Perfect numbers, which are just the same, and the form 6k - 2, except

    for the initial 6. Our demonstration Ref. 2).

    In detail, the table for the first 10 perfect numbers with pairs S and P:

    TAB. 3

    PERFECT NUMBER S P

    6 6 6

    28 1 7

    496 1 0

    8128 1 2

    33 550 336 1 0

    8 589 869 056 1 0

    137 438 691 328 1 0

    2 305 843 008 139 952 128 1 0

    2 658 455 991 569 831 744 654 692 615 953 842 176 1 0

    191 561 942 608 236 107 294 793 378 084 303 638 130 997

    321 548 169 2161 0

    The sum S = 1 because all perfect numbers are derived from the Mersenne numbers:

    The perfect number is given by 2p1

    (2p1) with p prime number but all numbers of the

    form 2m1

    (2m1) withm odd integer give the sum to S = 1.

    In fact, for the only perfect number 6 does not work because p = 2 and it is not an odd

    integer.

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    Moreover, all the perfect numbers end with digit 6 or 8.

    By

    2n

    (2n+1

    - 1)

    We have:

    2n

    is even and ends with digit 2, 4, 8, 6

    (2

    n+1

    - 1) is odd and ends with digit 3, 7, 5, 1.

    The value '5 'should be discarded as the hypothesis would fall primality, then the pairs

    that remain are (2,3), (4,7) and (6,1), whose products give the numbers 6 and 8, last

    digit of each perfect number.

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    2. CONCLUSIONS

    There is NO probabilistic primality test secure and efficient way to determine whether a

    number is prime or not.

    The only certainty that we have is the following:

    CONDITIONS NECESSARY BUT NOT SUFFICIENT to determine if a number N

    is prime is that his last digit is equal to 1, 3, 7 or 9 and that the sum of its digits S is

    not equal to 0, 3 or 6 because otherwise the number N is divisible by 3.

    Equivalently we have that the only universal regularity that emerges is that all prime

    numbers except 2 and 3 initials, are of the form 6k - 1 and 6k +1, since all otherpossible forms (6k-4, 6k-3. 6k-2, 6k) formed from composite numbers.

    There are no other conditions and assumptions.

    The sequence of prime numbers is intrinsically linked to increase to infinity of natural

    numbers N+ and is completely devoid of any rule, it is precisely this "no rule" thatdistributes random prime numbers.

    Even the famous Riemann hypothesis, surmising that it is true (it is yet to be proven)

    does not lead to any connection of prime numbers with the placement of the zeros of the

    Riemann zeta function.

    Although the Real part of the zeros always worth , the Imaginary parts are of no help

    to determine whether a number is prime or not.

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    In addition, the zeta function, useless as a primality test as discussed above, it is useless

    even to give aid to the factorization, for example that of the RSA numbers, as is oftenerroneously believed.

    Some few algorithm assumes RH in truth there, but is not dangerous enough, in fact

    encryption is RSA is still untouched. The zeta function does not give a list of prime

    numbers, nor a list of pairs of prime numbers useful for this purpose. Lists of the pairs

    give the twin primes conjecture, Goldbach's conjecture, Sophie Germain conjecture.

    These conjectures are ongoing demonstration, with our contributions also present.

    The products of two numbers twins of the same couple (or close to each other, ie withdifference 4, 6 etc..) are recognizable from the decimal part of their square root, veryhigh, eg. 0.999 ... and this type of product is easily factored by Fermat's algorithm used

    in reverse, ie from the square root of N = p * q with p and q twins or very close. It is

    located p immediately close to the square root of N (see our work on the assumption

    percentage)

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    3. ANALYSIS OF 1229 PRIMES LESS THAN 10000

    In Table 4 the complete list of pairs S and P 1229 for all prime numbers that are in the

    range 2 ... 10000

    TAB. 4

    N Prime Number S P1 2 2 2

    2 3 3 3

    3 5 5 5

    4 7 7 7

    5 11 2 1

    6 13 4 3

    7 17 8 7

    8 19 1 0

    9 23 5 610 29 2 0

    11 31 4 3

    12 37 1 3

    13 41 5 4

    14 43 7 3

    15 47 2 1

    16 53 8 6

    17 59 5 0

    18 61 7 619 67 4 6

    20 71 8 7

    21 73 1 3

    22 79 7 0

    23 83 2 6

    24 89 8 0

    25 97 7 0

    26 101 2 0

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    27 103 4 028 107 8 0

    29 109 1 0

    30 113 5 3

    31 127 1 5

    32 131 5 3

    33 137 2 3

    34 139 4 0

    35 149 5 0

    36 151 7 537 157 4 8

    38 163 1 0

    39 167 5 6

    40 173 2 3

    41 179 8 0

    42 181 1 8

    43 191 2 0

    44 193 4 0

    45 197 8 046 199 1 0

    47 211 4 2

    48 223 7 3

    49 227 2 1

    50 229 4 0

    51 233 8 0

    52 239 5 0

    53 241 7 8

    54 251 8 155 257 5 7

    56 263 2 0

    57 269 8 0

    58 271 1 5

    59 277 7 8

    60 281 2 7

    61 283 4 3

    62 293 5 0

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    99 523 1 3100 541 1 2

    101 547 7 5

    102 557 8 4

    103 563 5 0

    104 569 2 0

    105 571 4 8

    106 577 1 2

    107 587 2 1

    108 593 8 0109 599 5 0

    110 601 7 0

    111 607 4 0

    112 613 1 0

    113 617 5 6

    114 619 7 0

    115 631 1 0

    116 641 2 6

    117 643 4 0118 647 8 6

    119 653 5 0

    120 659 2 0

    121 661 4 0

    122 673 7 0

    123 677 2 6

    124 683 8 0

    125 691 7 0

    126 701 8 0127 709 7 0

    128 719 8 0

    129 727 7 8

    130 733 4 0

    131 739 1 0

    132 743 5 3

    133 751 4 8

    134 757 1 2

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    135 761 5 6136 769 4 0

    137 773 8 3

    138 787 4 5

    139 797 5 0

    140 809 8 0

    141 811 1 8

    142 821 2 7

    143 823 4 3

    144 827 8 4145 829 1 0

    146 839 2 0

    147 853 7 3

    148 857 2 1

    149 859 4 0

    150 863 8 0

    151 877 4 5

    152 881 8 1

    153 883 1 3154 887 5 7

    155 907 7 0

    156 911 2 0

    157 919 1 0

    158 929 2 0

    159 937 1 0

    160 941 5 0

    161 947 2 0

    162 953 8 0163 967 4 0

    164 971 8 0

    165 977 5 0

    166 983 2 0

    167 991 1 0

    168 997 7 0

    169 1009 1 0

    170 1013 5 0

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    171 1019 2 0172 1021 4 0

    173 1031 5 0

    174 1033 7 0

    175 1039 4 0

    176 1049 5 0

    177 1051 7 0

    178 1061 8 0

    179 1063 1 0

    180 1069 7 0181 1087 7 0

    182 1091 2 0

    183 1093 4 0

    184 1097 8 0

    185 1103 5 0

    186 1109 2 0

    187 1117 1 7

    188 1123 7 6

    189 1129 4 0190 1151 8 5

    191 1153 1 6

    192 1163 2 0

    193 1171 1 7

    194 1181 2 8

    195 1187 8 2

    196 1193 5 0

    197 1201 4 0

    198 1213 7 6199 1217 2 5

    200 1223 8 3

    201 1229 5 0

    202 1231 7 6

    203 1237 4 6

    204 1249 7 0

    205 1259 8 0

    206 1277 8 8

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    207 1279 1 0208 1283 5 3

    209 1289 2 0

    210 1291 4 0

    211 1297 1 0

    212 1301 5 0

    213 1303 7 0

    214 1307 2 0

    215 1319 5 0

    216 1321 7 6217 1327 4 6

    218 1361 2 0

    219 1367 8 0

    220 1373 5 0

    221 1381 4 6

    222 1399 4 0

    223 1409 5 0

    224 1423 1 6

    225 1427 5 2226 1429 7 0

    227 1433 2 0

    228 1439 8 0

    229 1447 7 4

    230 1451 2 2

    231 1453 4 6

    232 1459 1 0

    233 1471 4 1

    234 1481 5 5235 1483 7 6

    236 1487 2 8

    237 1489 4 0

    238 1493 8 0

    239 1499 5 0

    240 1511 8 5

    241 1523 2 3

    242 1531 1 6

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    243 1543 4 6244 1549 1 0

    245 1553 5 3

    246 1559 2 0

    247 1567 1 3

    248 1571 5 8

    249 1579 4 0

    250 1583 8 3

    251 1597 4 0

    252 1601 8 0253 1607 5 0

    254 1609 7 0

    255 1613 2 0

    256 1619 8 0

    257 1621 1 3

    258 1627 7 3

    259 1637 8 0

    260 1657 1 3

    261 1663 7 0262 1667 2 0

    263 1669 4 0

    264 1693 1 0

    265 1697 5 0

    266 1699 7 0

    267 1709 8 0

    268 1721 2 5

    269 1723 4 6

    270 1733 5 0271 1741 4 1

    272 1747 1 7

    273 1753 7 6

    274 1759 4 0

    275 1777 4 1

    276 1783 1 6

    277 1787 5 5

    278 1789 7 0

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    279 1801 1 0280 1811 2 8

    281 1823 5 3

    282 1831 4 6

    283 1847 2 8

    284 1861 7 3

    285 1867 4 3

    286 1871 8 2

    287 1873 1 6

    288 1877 5 5289 1879 7 0

    290 1889 8 0

    291 1901 2 0

    292 1907 8 0

    293 1913 5 0

    294 1931 5 0

    295 1933 7 0

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    OBSERVATIONS

    We note the frequent series of consecutive primes with P = 0

    Examples: (in red those close to a power of 10, except 10^1)

    From 89 a 109, with 29-23 = 6 consecutive prime numbers containing the digit 9

    From 349 to 419, with 81-69 = 12 consecutive numbers, starting with the digits 33 and3*3 =9, with P = 0

    From 907 a 1109, with 186 -154 = 32 consecutive primes, starting or containing the

    digit 9

    From 1879 to 2099, with 317-288 = 29 prime numbers containing a digit 9

    Another group of consecutive prime numbers with P = 0 is located between 2897 and

    3121, with 444-418 = 26 prime numbers, which also include the figure 9.

    The most numerous series are close to a power of 10 (ie, a bit before and a little 'later,like 6 prime numbers around 100).

    32 around in 1000 and 122 before 10,000 since 1129-1107 = 122 consecutive prime

    numbers, starting of course with 8, but contain the digit 9, or starting with 9. An 'other

    small series after 10 000 will complete the series for the fourth power of 10 = 10 ^ 4.

    Any number containing the digit 9, with regard to P = 0, multiplied by all the other

    digits of the number to be always a multiple of 9, so being modulo 9 = 0, its product isalways P = 0.

    The same is true also for the numbers, prime or not prime, containing two-digit 3, also

    not consecutive, because 3 * 3 = 9, and then all the products with two 3 digits in the

    factors are congruent modulo 9 and then P = 0, and so also the digits 3 and 6, since 6 * 3

    = 18 multiple of 9, and then the even its products with other digits of the number, and

    hence P = 0 in all these cases.

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    Let us now see a table with S and P related to prime numbers and composite of form 6k

    -1 and 6k +1

    Table 6

    6k -1 primes andcompositeS P

    6k+1 primes andcomposite

    Observations for the pairs(S, P) for compositenumbers

    55 5

    71 7

    112 1

    134 3

    178 7

    191 0

    235 6

    257 1 7 1

    29

    11 2

    31

    4 3358 6

    371 3 8 6

    415 4

    437 3

    472 1

    494 0 4 0

    538 6

    551 7 1 7

    595 0

    617 6

    652 3

    67

    718 7

    731 3

    775 0

    797 0 5 0

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    Last column:

    7 1

    8 6

    4 0

    1 7

    5 0

    1 0

    5 0

    7 5

    2 0

    8 1

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    7 1 exists only once8 6 exists only once4 0 exists only once1 7 exists only once5 0 exists only once1 0 exists only once7 5 exists only once

    2 0 exists only once8 1 exists only once

    Conclusions:

    If a number N of such forms 6k-1 and 6k + 1 has one of these pairs, it is probably

    composite. For example, we see the multiple of 7, then 7 * p

    Table 7

    7*p Pairs S and P Observations

    35 8 6

    49 4 0

    77 5 0

    91 1 0

    119 2 0

    161 8 6

    203 5 0

    217 1 5

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    Pairs for numbers N = 7*p

    8 6

    4 0

    5 0

    1 0

    2 0

    8 6

    repetition4 0repetition

    List of pairs for composite numbers (from Table 6)

    7 1 exists only once8 6 exists only once

    4 0 exists only once1 7 exists only once5 0 there is twice1 0 exists only once7 5 exists only once2 0 exists only once8 1 exists only once

    The two lists match, although in a different order, in the second list is missing 1 5,which appears for the first time for

    N = 7*31 =217

    Same thing could be done to the multiple of 11, up to p = 31

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    11*p Pairs S and P Observations

    55 1 7

    77 5 0

    121 4 2

    143 8 3

    187 7 2

    209 2 0

    253 1 3

    341 8 3

    Some couples (in brown) are repeated, others appear new (in purple)

    Pairs (S, P)

    1 7

    5 0

    4 2

    8 37 2

    2 0

    1 3

    8 3

    The pairs (1, 7), (5, 0) and (2, 0) repeat for 11 * p, so there may be repeated for some of

    the next multiple, and therefore are not evidence of their primality.

    For example, extending the table to 11 * p, we have

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    11*p Pairs (S, P) Observations

    55 1 7

    77 5 0

    121 4 2

    143 8 3

    187 7 2

    209 2 0

    253 1 3

    341 (= 11*31) 8 3repetition

    Seconda parte

    407 (=11*39) 2 0repetition

    451 1 2

    473 5 3

    517 4 8

    561 3 0

    737 8 3repetition

    781 7 6

    803 (= 11*73) 2 0repetition

    869 (=11*79) 5 0repetition

    Pairs first part

    Pairs (S, P

    1 7

    5 0

    4 2

    8 3

    7 2

    2 0

    1 3

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    Pairs second part

    2 0repetition

    1 2

    5 3

    4 8

    3 08 3repetition

    7 6

    2 0repetition

    5 0repetition

    (5. 0), (8, 3), (2, 0), are the pairs (S, P) most frequent for numbers N multiple of 11. Forexample, if we have the number 1067 to be tested, we find the pair (5, 0), the second of

    this list, and then 1067 may be not a prime number, and a multiple of 11, in fact 1067 =11 * 97. Additional tables for multiple other primes might allow to identify their pairs

    (S, P) "favorite", and testing them, we would find that often are not prime numbers, but

    particular multiple of a prime number (11 in this case) Another example:

    N = 1397,

    a pair (2, 0), present a favorite of 11 * p

    In fact, 1397 = 11 * 127.

    These are useful, but not deterministic primality test or not primality test.

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    CONCLUSIONS

    On this work we have discovered something about the primality test, which may be

    explored in the future with both our or others further research.

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    4. REFERENCES

    1a) Test di primalit AKS, da Wikipedia :

    Algoritmo AKS

    Da Wikipedia, l'enciclopedia libera.

    Questa voce sull'argomento teoria dei numeri solo un abbozzo.

    Contribuisci a migliorarla secondo le convenzioni di Wikipedia.

    L'algoritmo AKS (dalle iniziali dei tre ideatori, i matematici indiani Manindra Agrawal,Neeraj Kayal e Nitin Saxena) un test di primalit di complessit polinomiale. In

    particolare, l'algoritmo ha tempo di esecuzione O(log12+

    n), mentre una variante

    proposta nel 2005 da Carl Pomerance e Hendrik Lenstra ha complessit O(log6+

    n)

    Pubblicato nel 2002, ha fruttato ai suoi scopritori diversi premi, tra cui il premio Gdel

    e il premio Fulkerson nel 2006.

    1b) AKS primality test (in inglese)From Wikipedia, the free encyclopedia

    Jump to: navigation, search

    The AKS primality test (also known as AgrawalKayalSaxena primality test and

    cyclotomic AKS test) is a deterministic primality-proving algorithm created and

    published by Manindra Agrawal, Neeraj Kayal, and Nitin Saxena, computer scientists at

    the Indian Institute of Technology Kanpur, on August 6, 2002, in a paper titled

    "PRIMES is in P".[1]

    The algorithm determines whether a number is prime or composite

    within polynomial time. The authors received the 2006 Gdel Prize and the 2006

    Fulkerson Prize for this work.Contents

    1 Importance

    2 Concepts

    3 History and running time4 Algorithm

    5 References

    6 Further reading

    7 External links

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    ImportanceAKS is the first primality-proving algorithm to be simultaneously general, polynomial,

    deterministic, and unconditional. Previous algorithms had been developed for centuriesbut achieved three of these properties at most, but not all four.

    The AKS algorithm can be used to verify the primality of any general number given.

    Many fast primality tests are known that work only for numbers with certain properties.

    For example, the LucasLehmer test for Mersenne numbers works only for Mersenne

    numbers, while Ppin's test can be applied to Fermat numbers only.

    The maximum running time of the algorithm can be expressed as a polynomial over the

    number of digits in the target number. ECPP and APR conclusively prove or disprovethat a given number is prime, but are not known to have polynomial time bounds for allinputs.

    The algorithm is guaranteed to distinguish deterministically whether the target numberis prime or composite. Randomized tests, such as MillerRabin and BailliePSW, can

    test any given number for primality in polynomial time, but are known to produce only

    a probabilistic result.

    The correctness of AKS is not conditional on any subsidiary unproven hypothesis. In

    contrast, the Miller test is fully deterministic and runs in polynomial time over all

    inputs, but its correctness depends on the truth of the yet-unproven generalized Riemannhypothesis.

    Concepts

    The AKS primality test is based upon the following theorem: An integer n ( 2) is

    prime if and only if the polynomial congruence relation

    holds for all integers a coprime to n (or even just for some such integer a, in particular

    for a = 1).[1]

    Note that x is a free variable. It is never substituted by a number; instead

    you have to expand and compare the coefficients of the x powers.

    This theorem is a generalization to polynomials of Fermat's little theorem, and caneasily be proven using the binomial theorem together with the following property of the

    binomial coefficient:

    for all if and only if n is prime.

    While the relation (1) constitutes a primality test in itself, verifying it takes exponential

    time. Therefore, to reduce the computational complexity, AKS makes use of the related

    congruence

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    which is the same as:

    for some polynomials f and g. This congruence can be checked in polynomial time withrespect to the number of digits in n, because it is provable that r need only be

    logarithmic with respect to n. Note that all primes satisfy this relation (choosing g = 0 in

    (3) gives (1), which holds for n prime). However, some composite numbers also satisfy

    the relation. The proof of correctness for AKS consists of showing that there exists a

    suitably small r and suitably small set of integers A such that, if the congruence holds

    for all such a in A, then n must be prime.

    2) I NUMERI PERFETTI DISPARI

    (proposta di dimostrazione della loro inesistenza)

    Gruppo B. Riemann

    Michele Nardelli, Francesco Di Noto

    Aggiungiamo un generatore di numeri primi creato da

    (Marco Cavicchioli) e pubblicato sul sito

    (http://www.MarcoCavicchioli.it/numeri_primi_generatore.html

    3) Generatore di numeri primi in Javascript

    di Marco Cavicchioli

    Inserire un numero intero positivo (N) maggiore di 1 nel seguente campo e cliccare su

    "Genera":Inizio modulo

    N =79

    Fine modulo

    Una volta cliccato su "Genera" un codice Javascript estrarr dalla successione dei

    numeri interi positivi minori o uguali a N l'elenco dei numeri primi in essa riconosciuti.

    Dato che la funzione Javascript che esegue ci genera (n-1) successioni da (n-1) a (n-1)2

    la sua esecuzione risulta molto lenta per N molto alti (per esempio oltre 1000).

    Da questo elenco convenzionalmente escluso il numero 1, perch tutti i numeri interi

    positivi sono multipli di 1.Per vedere il codice Javascript sufficiente aprire questa pagina html con un editor di

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    testo. Per tenere questo codice il pi pulito possibile ho scelto di non includere in questapagina alcun elemento grafico.

    Ovviamente tutto il materiale contenuto in questa pagina (sia testo che codice) CopyLeft, ovvero disponibile liberamente a chiunque. L'autore tuttavia richiede

    espressamente che, qualora venisse utilizzato, citato o ripubblicato, in ogni luogo e

    forma, vengano sempre riportati con chiarezza ed evidenza il nome dell'autore (Marco

    Cavicchioli) e l'url (http://www.MarcoCavicchioli.it/numeri_primi_generatore.html).

    Questa pagina ottimizzata per Mozilla Firefox.

    UPDATE: segnalo che stato pubblicato un TEOREMA FONDAMENTALE DEI

    NUMERI PRIMI che merita di essere letto.

    Descrizione della funzione

    I numeri interi positivi maggiori di 1 sarebbero in realt tutti numeri multipli (di 1

    stesso, che il primo numero intero positivo, e quello da cui vengono generati per

    somma tutti gli altri numeri interi positivi). Per "numero primo" invece si intende un

    numero che non multiplo (ovvero: un numero intero positivo o multiplo, o primo). Faccio notare che, a rigor di logica, a questo punto 1 sarebbe l'unico numeroprimo, perch l'unico a non essere multiplo (dato che tutti gli altri numeri interi positivi

    sono perlomeno multipli di 1). Ma questo non esclude che si possa procedere lo stesso,con un'apposita accortezza...

    Estrarre la successione dei numeri primi minori o uguali a N pertanto significa estrarre

    dalla successione di tutti i numeri interi positivi minori o uguali a N quelli che non siano

    soltanto multipli di 1. Questo concetto presume che si escluda la possibilit che un

    numero N possa essere multiplo di s stesso. Per tentare di convincervi di ci devo fare

    un passo indietro.

    Se accettiamo il presupposto che tutti i numeri interi positivi siano in realt funzioni di 1

    (ovvero valori risultanti dall'applicazione di una data funzione somma al numero 1,come se in realt esistessero solo due numeri "primordiali", 0 e 1, con tutti i restanti

    numeri interi superiori in realt ricavati da somme di 1, e pertanto "multipli di 1") allora

    tutti i numeri interi positivi, con l'esclusione di 1, sarebbero somme di "uni" (es. 2=1+1,

    3=1+1+1, 7=1+1+1+1+1+1+1, ecc.). A questo punto potremmo definire "multipli"

    (sempre con l'esclusione di 1 stesso) tutti i numeri interi positivi che si ottengono

    sommando ripetutamente dei "pattern di uni". Per pattern di uni intendo sequenze

    identiche ripetute di n uni (ad esempio (1+1+1)+(1+1+1)+(1+1+1)=9 dimostra che 9

    un numero multiplo). Se imponiamo che, in questo caso, n sia superiore a 1 allora

    otteniamo che i numeri primi sono quei numeri che non si ottengono da sommeripetute di pattern di n uni, con n>1. Il fatto che sia richiesto che tali somme sano

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    ripetute esclude che un numero intero positivo N possa essere multiplo di s stesso,perch in questo caso il pattern di uni non potrebbe essere ripetutamente sommato per

    ottenere N, se n vale N.Pertanto posso ora elencare alcune considerazioni importanti:

    i numeri multipli sono quei numeri interi positivi che si ottengono da somme ripetute di

    pattern di n uni, con n>1

    dato che tali somme di pattern devono essere ripetute, un numero intero positivo N non

    pu essere multiplo di s stesso, perch n deve essere minore di N (n

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    4) REGOLA UNIVERSALE PER TROVARE TUTTI I NUMERI

    PRIMI

    Ing. Pier Francesco Roggero, Dott. Michele Nardelli, Francesco Di Noto

    gi sul nostro sito http://nardelli.xoom.it/virgiliowizard/