Are there any composite numbers in this equation?

heavyc · Apr 25, 2005

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

heavyc · Apr 25, 2005

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

Muzza · Apr 25, 2005

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).

heavyc · Apr 25, 2005

Muzza said:

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).

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?

Muzza · Apr 25, 2005

Well, I simply tried all possible factors (with a computer, obviously).

heavyc · Apr 25, 2005

Muzza said:

Well, I simply tried all possible factors (with a computer, obviously).

Well what's I mean how do you know that it is composite what steps do you do to prove that it is composite?

Muzza · Apr 25, 2005

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.

heavyc · Apr 25, 2005

Muzza said:

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.

I still don't understand by what you mean can you give me an example why 3233 is composite?

Muzza · Apr 25, 2005

3233 is composite because 3233 = 53 * 61.

heavyc · Apr 25, 2005

Muzza said:

3233 is composite because 3233 = 53 * 61.

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 appreciate it. You were very helpful

dextercioby · Apr 25, 2005

f(x)=x^2-79x+1601

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$f\left( 2\right) =\allowbreak 1447=\allowbreak 1447$ $f\left( 10\right) =\allowbreak 911=\allowbreak 911$ $f\left( 18\right) =\allowbreak 503=\allowbreak 503$ $f\left( 26\right) =\allowbreak 223=\allowbreak 223$ 038$f\left( 3\right) =\allowbreak 1373=\allowbreak 1373$ $f\left( 11\right) =\allowbreak 853=\allowbreak 853$ $f\left( 19\right) =\allowbreak 461=\allowbreak 461$ $f\left( 27\right) =\allowbreak 197=\allowbreak 197$ 038$f\left( 4\right) =\allowbreak 1301=\allowbreak 1301$ $f\left( 12\right) =\allowbreak 797=\allowbreak 797$ $f\left( 20\right) =\allowbreak 421=\allowbreak 421$ $f\left( 28\right) =\allowbreak 173=\allowbreak 173$ 038$f\left( 5\right) =\allowbreak 1231=\allowbreak 1231$ $f\left( 13\right) =\allowbreak 743=\allowbreak 743$ $f\left( 21\right) =\allowbreak 383=\allowbreak 383$ $f\left( 29\right) =\allowbreak 151=\allowbreak 151$ 038$f\left( 6\right) =\allowbreak 1163=\allowbreak 1163$ $f\left( 14\right) =\allowbreak 691=\allowbreak 691$ $f\left( 22\right) =\allowbreak 347=\allowbreak 347$ $f\left( 30\right) =\allowbreak 131=\allowbreak 131$ 038$f\left( 7\right) =\allowbreak 1097=\allowbreak 1097$ $f\left( 15\right) =\allowbreak 641=\allowbreak 641$ $f\left( 23\right) =\allowbreak 313=\allowbreak 313$ $f\left( 31\right) =\allowbreak 113=\allowbreak 113$ 038$f\left( 8\right) =\allowbreak 1033=\allowbreak 1033$ $f\left( 16\right) =\allowbreak 593=\allowbreak 593$ $f\left( 24\right) =\allowbreak 281=\allowbreak 281$ $f\left( 32\right) =\allowbreak 97=\allowbreak 97$ 038$f\left( 33\right) =\allowbreak 83=83$ $f\left( 79\right) =\allowbreak 1601=\allowbreak 1601$ $f\left( 87\right) =\allowbreak 2297=\allowbreak 2297$ $f\left( 95\right) =\allowbreak 3121=\allowbreak 3121$ 038$f\left( 34\right) =\allowbreak 71=71$ $f\left( 80\right) =\allowbreak 1681=\allowbreak 41^2$ $f\left( 88\right) =\allowbreak 2393=\allowbreak 2393$ $f\left( 96\right) =\allowbreak 3233=\allowbreak 53\times 61$ 038$f\left( 35\right) =\allowbreak 61=61$ $f\left( 81\right) =\allowbreak 1763=\allowbreak 41\times 43$ $f\left( 89\right) =\allowbreak 2491=\allowbreak 47\times 53$ $f\left( 97\right) =\allowbreak 3347\allowbreak =\allowbreak 3347$ 038$f\left( 36\right) =\allowbreak 53=53$ $f\left( 82\right) =\allowbreak 1847=\allowbreak 1847$ $f\left( 90\right) =\allowbreak 2591=\allowbreak 2591$ $f\left( 98\right) =\allowbreak 3463=\allowbreak 3463$ 038$f\left( 37\right) =\allowbreak 47=47$ $f\left( 83\right) =\allowbreak 1933=\allowbreak 1933$ $f\left( 91\right) =\allowbreak 2693=\allowbreak 2693$ $f\left( 99\right) =\allowbreak 3581=\allowbreak 3581$ 038$f\left( 38\right) =\allowbreak 43=43$ $f\left( 84\right) =\allowbreak 2021=\allowbreak 43\times 47$ $f\left( 92\right) =\allowbreak 2797=\allowbreak 2797$ $f\left( 100\right) =\allowbreak 3701=\allowbreak 3701$ 038$f\left( 39\right) =\allowbreak 41=41\,f\left( 85\right) $ $=2111=\allowbreak 2111$ $f\left( 93\right) =\allowbreak 2903=\allowbreak 2903$ 038$f\left( 40\right) =\allowbreak 41=41$ $f\left( 86\right) =\allowbreak 2203=\allowbreak 2203$ $f\left( 94\right) =\allowbreak 3011=\allowbreak 3011$ Daniel.

neurocomp2003 · Apr 25, 2005

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

OlderDan · Apr 25, 2005

heavyc said:

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

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

f(x)=x^2-79x+1601

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

f(x)=x^2-79x+1601 = (x+a)(x-a-79) =x^2-79x-a^2-79a

This has a solution only if

-a^2 - 79a =1601

a^2 + 79a + 1601 = 0

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

a = -\frac{79}{2}\pm\sqrt{-40.75}

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

53*61 = 3233 = 96^2 - 79*96 + 1601

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

a = 53 - 96 = -43

x + a = 53

x - a - 79 = 96 + 43 - 79 = 60

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

a = 61 - 96 = -35

x + a = 61

x - a - 79 = 96 + 35 - 79 = 52

Of course, I don't expct the second number to be correct because I found no solution for a in the first place.

Data · Apr 25, 2005

If you could find such an integer a, 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 x) every integer x.

Essentially, just because you can factor the polynomial that way for some particular values of x does not mean that you can do so for every x.

Are there any composite numbers in this equation?

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