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Are there any composite numbers in this equation?

  1. Apr 25, 2005 #1
    I was just wondering if there are any composite numbers in this equation
    for every number from k = 1 - 100
    k * k - 79 * k + 1601

    all the numbers that I have come across seem to be prime but I could be wrong so if some one could check my work Please it would be helpful
     
  2. jcsd
  3. Apr 25, 2005 #2
    here are all the numbers in the sequence!

    These are all the numbers in the sequence if someone could take a look at them and check me because i think that all the numbers are prime.

    [k] [number after equation]
    1 1523
    2 1447
    3 1373
    4 1301
    5 1231
    6 1163
    7 1097
    8 1033
    9 971
    10 911
    11 853
    12 797
    13 743
    14 691
    15 641
    16 593
    17 547
    18 503
    19 461
    20 421
    21 383
    22 347
    23 313
    24 281
    25 251
    26 223
    27 197
    28 173
    29 151
    30 131
    31 113
    32 97
    33 83
    34 71
    35 61
    36 53
    37 47
    38 43
    39 41
    40 41
    41 43
    42 47
    43 53
    44 61
    45 71
    46 83
    47 97
    48 113
    49 131
    50 151
    51 173
    52 197
    53 223
    54 251
    55 281
    56 313
    57 347
    58 383
    59 421
    60 461
    61 503
    62 547
    63 593
    64 641
    65 691
    66 743
    67 797
    68 853
    69 911
    70 971
    71 1033
    72 1097
    73 1163
    74 1231
    75 1301
    76 1373
    77 1447
    78 1523
    79 1601
    80 1681
    81 1763
    82 1847
    83 1933
    84 2021
    85 2111
    86 2203
    87 2297
    88 2393
    89 2491
    90 2591
    91 2693
    92 2797
    93 2903
    94 3011
    95 3121
    96 3233
    97 3347
    98 3463
    99 3581
    100 3701
     
  4. Apr 25, 2005 #3
    k * k - 79 * k + 1601 is composite for k = 80, 81, 84, 89 and 96, all the rest are prime (for 1 <= k <= 100, of course).
     
    Last edited: Apr 25, 2005
  5. Apr 25, 2005 #4
    how would you calculate them to be composite because I have a hard time figuring this out because I think that a prime number can only be divisible by 1 and composite can be divisible by more than 2 factors. Would you be able to clarify this for me????
     
  6. Apr 25, 2005 #5
    Well, I simply tried all possible factors (with a computer, obviously).
     
  7. Apr 25, 2005 #6
    Well whats I mean how do you know that it is composite what steps do you do to prove that it is composite???
     
  8. Apr 25, 2005 #7
    As I said, I simply tried all possible factors. If I found one that was non-trivial (i.e. a factor not equal to the number itself, or 1), I knew that the number was composite.
     
  9. Apr 25, 2005 #8
    I still dont understand by what you mean can you give me an example why 3233 is composite?
     
  10. Apr 25, 2005 #9
    3233 is composite because 3233 = 53 * 61.
     
  11. Apr 25, 2005 #10
    O ok i think I get it now so ur saying that if I take the number and try and multiply it by any numbers that I can and try to get that number? Thank You for the help I really appriciate it. You were very helpful
     
  12. Apr 25, 2005 #11

    dextercioby

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    Science Advisor
    Homework Helper

    [tex]f(x)=x^2-79x+1601 [/tex]

    [tex] f\left( 1\right) =\allowbreak 1523=\allowbreak 1523$
    $f\left( 9\right) =\allowbreak 971=\allowbreak 971$ $f\left( 17\right) =\allowbreak 547=\allowbreak 547$ $f\left( 25\right) =\allowbreak 251=\allowbreak 251$

    038<p type="texpara" tag="Body Text" >$f\left( 2\right) =\allowbreak 1447=\allowbreak 1447$ $f\left( 10\right) =\allowbreak 911=\allowbreak 911$ $f\left( 18\right) =\allowbreak 503=\allowbreak 503$ $f\left( 26\right) =\allowbreak 223=\allowbreak 223$

    038<p type="texpara" tag="Body Text" >$f\left( 3\right) =\allowbreak 1373=\allowbreak 1373$ $f\left( 11\right) =\allowbreak 853=\allowbreak 853$ $f\left( 19\right) =\allowbreak 461=\allowbreak 461$ $f\left( 27\right) =\allowbreak 197=\allowbreak 197$

    038<p type="texpara" tag="Body Text" >$f\left( 4\right) =\allowbreak 1301=\allowbreak 1301$ $f\left( 12\right) =\allowbreak 797=\allowbreak 797$ $f\left( 20\right) =\allowbreak 421=\allowbreak 421$ $f\left( 28\right) =\allowbreak 173=\allowbreak 173$

    038<p type="texpara" tag="Body Text" >$f\left( 5\right) =\allowbreak 1231=\allowbreak 1231$ $f\left( 13\right) =\allowbreak 743=\allowbreak 743$ $f\left( 21\right) =\allowbreak 383=\allowbreak 383$ $f\left( 29\right) =\allowbreak 151=\allowbreak 151$

    038<p type="texpara" tag="Body Text" >$f\left( 6\right) =\allowbreak 1163=\allowbreak 1163$ $f\left( 14\right) =\allowbreak 691=\allowbreak 691$ $f\left( 22\right) =\allowbreak 347=\allowbreak 347$ $f\left( 30\right) =\allowbreak 131=\allowbreak 131$

    038<p type="texpara" tag="Body Text" >$f\left( 7\right) =\allowbreak 1097=\allowbreak 1097$ $f\left( 15\right) =\allowbreak 641=\allowbreak 641$ $f\left( 23\right) =\allowbreak 313=\allowbreak 313$ $f\left( 31\right) =\allowbreak 113=\allowbreak 113$

    038<p type="texpara" tag="Body Text" >$f\left( 8\right) =\allowbreak 1033=\allowbreak 1033$ $f\left( 16\right) =\allowbreak 593=\allowbreak 593$ $f\left( 24\right) =\allowbreak 281=\allowbreak 281$ $f\left( 32\right) =\allowbreak 97=\allowbreak 97$

    038<p type="texpara" tag="Body Text" >$f\left( 33\right) =\allowbreak 83=83$ $f\left( 79\right) =\allowbreak 1601=\allowbreak 1601$ $f\left( 87\right) =\allowbreak 2297=\allowbreak 2297$ $f\left( 95\right) =\allowbreak 3121=\allowbreak 3121$

    038<p type="texpara" tag="Body Text" >$f\left( 34\right) =\allowbreak 71=71$ $f\left( 80\right) =\allowbreak 1681=\allowbreak 41^2$ $f\left( 88\right) =\allowbreak 2393=\allowbreak 2393$ $f\left( 96\right) =\allowbreak 3233=\allowbreak 53\times 61$

    038<p type="texpara" tag="Body Text" >$f\left( 35\right) =\allowbreak 61=61$ $f\left( 81\right) =\allowbreak 1763=\allowbreak 41\times 43$ $f\left( 89\right) =\allowbreak 2491=\allowbreak 47\times 53$ $f\left( 97\right) =\allowbreak 3347\allowbreak =\allowbreak 3347$

    038<p type="texpara" tag="Body Text" >$f\left( 36\right) =\allowbreak 53=53$ $f\left( 82\right) =\allowbreak 1847=\allowbreak 1847$ $f\left( 90\right) =\allowbreak 2591=\allowbreak 2591$ $f\left( 98\right) =\allowbreak 3463=\allowbreak 3463$

    038<p type="texpara" tag="Body Text" >$f\left( 37\right) =\allowbreak 47=47$ $f\left( 83\right) =\allowbreak 1933=\allowbreak 1933$ $f\left( 91\right) =\allowbreak 2693=\allowbreak 2693$ $f\left( 99\right) =\allowbreak 3581=\allowbreak 3581$

    038<p type="texpara" tag="Body Text" >$f\left( 38\right) =\allowbreak 43=43$ $f\left( 84\right) =\allowbreak 2021=\allowbreak 43\times 47$ $f\left( 92\right) =\allowbreak 2797=\allowbreak 2797$ $f\left( 100\right) =\allowbreak 3701=\allowbreak 3701$

    038<p type="texpara" tag="Body Text" >$f\left( 39\right) =\allowbreak 41=41\,f\left( 85\right) $ $=2111=\allowbreak 2111$ $f\left( 93\right) =\allowbreak 2903=\allowbreak 2903$

    038<p type="texpara" tag="Body Text" >$f\left( 40\right) =\allowbreak 41=41$ $f\left( 86\right) =\allowbreak 2203=\allowbreak 2203$ $f\left( 94\right) =\allowbreak 3011=\allowbreak 3011$

    Daniel.
     
    Last edited: Apr 25, 2005
  13. Apr 25, 2005 #12
    code it in C or Matlab.
    iterate over K
    get the number N = equation
    iterate over prime factors up to the sqrt N
    see if those prime factor divide N
    if they don't ....then N is prime
     
  14. Apr 25, 2005 #13

    OlderDan

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    Science Advisor
    Homework Helper

    Why doesn't this work? There has to be some nonsense in here, but so far I cannot find it.

    [tex]f(x)=x^2-79x+1601 [/tex]

    Assume f(x) has integer factors for some integer values of x. Let one factor be x + a where a is some integer, and divide the quadratic by x + a to get the other factor

    [tex]f(x)=x^2-79x+1601 = (x+a)(x-a-79) =x^2-79x-a^2-79a[/tex]

    This has a solution only if

    [tex]-a^2 - 79a =1601 [/tex]

    [tex]a^2 + 79a + 1601 = 0[/tex]

    [tex]a = -\frac{79}{2}\pm\sqrt{\left[\frac{79}{2}\right]^2-1601}[/tex]

    [tex]a = -\frac{79}{2}\pm\sqrt{-40.75}[/tex]

    This has no real solutions. It's bad enough that it has no solutions, but even if it did it would be independent of x, which seems unlikely. Where did I go wrong?

    From empirical evidence we know that

    [tex]53*61 = 3233 = 96^2 - 79*96 + 1601[/tex]

    If I set the first factor to x + a, I get

    [tex]a = 53 - 96 = -43[/tex]

    [tex]x + a = 53[/tex]

    [tex]x - a - 79 = 96 + 43 - 79 = 60[/tex]

    If I set the seond factor to x + a, I get

    [tex]a = 61 - 96 = -35[/tex]

    [tex]x + a = 61[/tex]

    [tex]x - a - 79 = 96 + 35 - 79 = 52[/tex]

    Of course, I don't expct the second number to be correct because I found no solution for a in the first place.
     
  15. Apr 25, 2005 #14
    If you could find such an integer [itex]a[/itex], it would mean that the polynomial had a composite integer value for (almost; specifically, it would take prime values at most for four values of [itex]x[/itex]) every integer [itex]x[/itex].

    Essentially, just because you can factor the polynomial that way for some particular values of [itex]x[/itex] does not mean that you can do so for every [itex]x[/itex].
     
    Last edited: Apr 25, 2005
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