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I Compositeness test

  1. Sep 11, 2016 #1
    All composite number comprised of atleast two factor e.g

    57 = 3 * 19 and 1 * 57
    14927= 11*1357 = 23*649 = 59*253 = 1*14927

    We can get to know 57 is a composite number , as the factor 3*19 is achieved, as below, from 11^2. The first factor reduces and the 2nd factor increases, till first factor is 1.


    Note all these composite no. that descend from 11^2 are square distance from 11^2 as below.

    121-117=4= 2^2
    121-105=16 = 4^2
    121-85=36 = 6^2
    121-57=64 = 8^2
    121-21= 100 = 10^2

    Note 1*21 factor has always a difference of 1, here, 11-10=1 (i.e of 11^2 & 10^2). So it can always be eliminated.

    This is true for all composite number. So, each Composite number has a specific square(higher than the test number N) from where it descend, like in the above example.

    So to find whether a number N is a composite or not. One only need to see if any of the higher squares are square distance to N.

    Say a test number N=1656337 . Square next to it is , integer(√N)+ 1

    which is,
    = integer (√1656337) ) + 1 = integer ( 1286.987...) ) + 1 = 1286 + 1 = 1287.

    So, square next to N is 1287^2.

    We can formulate distances to next squares which will be quadratic in nature. For the above number N, it will be,

    x^2 + 1287*2*x + (1287^2-N)

    (this is the distances to all the higher square to N. Starting with x=0)

    x^2 + 2574x + 32 = m^2 ???

    So, if the above quadratic expression generates an perfect square (integer). Then N is a composite number. Remember 1*N also comes under a specific squares, so every such equation will definitely have one solution for 1*N factor.

    So we need to see if the quadratic expression generates atleast two solution. If yes then definitely N is a composite number.

    But I am not able to determine, how to say such a quadratic will generate perfect squares or not.

    For the above test number N. Since I know 1656337 is a composite number and 1217 and 1361 are its factor.

    In that case, we can find the squares where its distance comes to be a square.
    (1217 + 1361) /2 = 1289

    So we get from 1289^2, 1217*1361 has descended.

    so, 1289^2 - 1656337 = 72^2


    x^2 + 2574x + 32 = 72^2
    so we definitely get at x=2 the quadratic is a perfect square

    Similarly for factor 1*1656337, it always descend from

    integer(1656337/2) + 1 = integer(828168.5)+1 = 828168+1 = 828169 square

    x^2 + 2574x + 32 = 828169^2
    and x for this, will always be

    x= [(1287-1)^2 - 32 ] / 2 = 826882

    x^2 + 2574x + 32 = m^2
    has two solution.

    i.e N(1656337) has two square at x= 2 & x = 826882 and therefore is a composite number.

    Please comment.
    Thank you.
  2. jcsd
  3. Sep 11, 2016 #2


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    2017 Award

    Staff: Mentor

    Try to find a square for 6=2*3. It does not work for even numbers (although it is trivial to find a prime factor for them).
    This is much more work than the usual tests if a number is prime.

    With a good algorithm, your home computer can check if a 50-digit number is prime in seconds, and find the factors in seconds to minutes. Example script.
    = 132985692076982233761143 * 34323499582363159679580258647 - found in 2 seconds.

    With the square numbers, it would have to run for years to millenia. Your approach doesn't look wrong, but it does not help.
  4. Sep 11, 2016 #3
    x^2 + 2574x + 32 = m^2 ???
    Is it really, difficult (or no way), to determine ,if quadratic like above, can generate perfect square or not. I had an impression, it can be solved just like any other equation.
  5. Sep 11, 2016 #4


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    Staff: Mentor

    Try the example of 4564534346464575453645453857398573985358375558353521.

    It is not sufficient to find any larger square, you have to express the number as difference of two squares. And finding that via trial and error is not feasible - it is equivalent to check all possible factors. Squares are rare if the numbers are that large.
  6. Sep 11, 2016 #5
    Why do you say to solve x^2 + 2574x + 32 = m^2 ???
    it has to be via trial and error.
  7. Sep 11, 2016 #6


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    Staff: Mentor

    Despite centuries of searches, no one found a method that does not involve a lot of trial and error (for large numbers), or uses the opposite direction (find the primes first, then reconstruct the difference of squares if you are interested in it).
    Such a method would revolutionize cryptography. It would also come with a reasonable prize money (that one expired but I think there are follow-up prizes).

    It is easy with 4-digit numbers such as yours, because it involves just a few thousand attempts in the worst case. A computer can do that in a millisecond. A computer cannot do 1050 attempts - and that is the range where prime factorization gets interesting.
  8. Sep 12, 2016 #7
    Come on, I have solved various questions of this type for you already. We won't keep spoonfeeding you the answer. So you should start by showing us some work.
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