MHB Solving for n: 4 Divisors & n+1 Formula

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The discussion focuses on finding natural numbers n that have exactly four positive divisors, including 1 and n. It specifies that n+1 must equal four times the sum of the other two divisors. Participants explore the implications of these conditions and work through potential solutions. The problem emphasizes the relationship between the divisors and the formula involving n+1. Ultimately, the goal is to identify all such natural numbers n that satisfy these criteria.
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we are given :
(1)n $\in N$
(2) n has exactly 4 positive divisors (including 1 and n)
(3)n+1 is four times the sum of the other two divisors
please find all n
 
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Albert said:
we are given :
(1)n $\in N$
(2) n has exactly 4 positive divisors (including 1 and n)
(3)n+1 is four times the sum of the other two divisors
please find all n

Let other 2 factors be a,b. without loss of generality a < b Note that a and b both must be prime otherwise >2 other divisors or > 4 divisors
from given condition
$n= ab$
$n+1 = 4(a+b)$
or $ab +1 = 4a + 4b$
or $ab-4a - 4b + 1 = 0$
or $(a-4)(b-4) = 15= 1 * 15= 3 * 5$

$(a-4) = 1 , b = 4 = 15=> a = 5, b= 19=>n= 95$
$a-4 = 3 , b-4 = 5 => b= 9$ hence b is not prime
so only solution
$n = 95$
 
kaliprasad said:
Let other 2 factors be a,b. without loss of generality a < b Note that a and b both must be prime otherwise >2 other divisors or > 4 divisors
from given condition
$n= ab$
$n+1 = 4(a+b)$
or $ab +1 = 4a + 4b$
or $ab-4a - 4b + 1 = 0$
or $(a-4)(b-4) = 15= 1 * 15= 3 * 5$

$(a-4) = 1 , b = 4 = 15=> a = 5, b= 19=>n= 95$
$a-4 = 3 , b-4 = 5 => b= 9$ hence b is not prime
so only solution
$n = 95$
yes , you are right
 
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