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Please give me a detailed solution for the question

Let n > 18 be a positive integer such that n-1 and n+1 both are prime .

Prove that n has at least 8 different positive factors .

 
Darkside  Nov 10, 2018
 #1
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https://web2.0calc.com/questions/help-me_137

 
Guest Nov 10, 2018
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60 =22 x 3 x 5 =3 "Prime Factors"
Total divisors =The exponent of each "Prime Factor" + 1, all multiplied together: (2+1) x (1+1)x(1+1) =12 divisors:
1 | 2 | 3 | 4 | 5 | 6 | 10 | 12 | 15 | 20 | 30 | 60 (12 divisors), or factors.
60 - 1 =59 Prime number
60+1 =61 Prime number.

 
Guest Nov 10, 2018
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deleted

 
Guest Nov 10, 2018
edited by Guest  Nov 10, 2018
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35802 =2 x 34 x 13 x 17 =4 "Prime Factors"
Total divisors =The exponent of each "Prime Factor" + 1, all multiplied together: (1+1) x (4+1)x(1+1)x(1+1) =40 divisors:
1 | 2 | 3 | 6 | 9 | 13 | 17 | 18 | 26 | 27 | 34 | 39 | 51 | 54 | 78 | 81 | 102 | 117 | 153 | 162 | 221 | 234 | 306 | 351 | 442 | 459 | 663 | 702 | 918 | 1053 | 1326 | 1377 | 1989 | 2106 | 2754 | 3978 | 5967 | 11934 | 17901 | 35802 (40 divisors), or factors
35802 - 1 =35801 Prime number
35802+1 =35803 Prime number.

 
Guest Nov 10, 2018
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I would also like to see this proof.  frown

 
Melody  Nov 10, 2018
edited by Melody  Nov 10, 2018
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This question applies only to those composite numbers that are between 2 "Twin Primes" that are >18. Here is a list of twin primes >18 and <1,000.    (29, 31) | (41, 43) | (59, 61) | (71, 73) | (101, 103) | (107, 109) | (137, 139) | (149, 151) | (179, 181) | (191, 193) | (197, 199) | (227, 229) | (239, 241) | (269, 271) | (281, 283) | (311, 313) | (347, 349) | (419, 421) | (431, 433) | (461, 463) | (521, 523) | (569, 571) | (599, 601) | (617, 619) | (641, 643) | (659, 661) | (809, 811) | (821, 823) | (827, 829) | (857, 859) | (881, 883) ...
    The definition of a prime number is that it has ONLY 2 prime divisors, i.e., itself and 1. Since two prime divisors will only have 2 exponents, which would only give:(1+1) x (1+1)=4 divisors. A composite number between 2 twin primes must, by definition,  necessarily have at least (1+1) extra divisor for a total of: (1+1) x (1+1) x (1+1) =8 divisors as " a minimum" to distinguish it from a prime number which MUST have only 4 divisors. And that is the best that I can give you as a "proof".

 
Guest Nov 11, 2018
edited by Guest  Nov 11, 2018
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This is incorrect, all prime numbers have exactly 2 divisors, not 4 (the prime number and 1), and not all composite numbers have 8 divisors.

 
Guest Nov 11, 2018
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Thanks guest. 1 is not a prime number but that is not central to your arguement.

I am afraid your second paragraph makes no sense to me at all.

Thanks for you input though :)

 
Melody  Nov 12, 2018

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