Solving for n: 4 Divisors & n+1 Formula

[Albert1]

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

 

[kaliprasad]

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

Spoiler

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
$n = 95$

 

[Albert1]

Spoiler

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