let Sn=n2+20n+12,n is a positive integer.what is the value of all possible values of n

Question

let S_n=n²+20n+12,n is a positive integer.what is the value of all possible values of n

Answer 1

 

Let y² = n² + 20n +12         

Since S_n has to be a perfect square, y is an integer 

 

y² + 88 = n² + 20n +100
y² + 88 = (n +10)²

 

 n + 10 = x  

(since n is an positive integer x will be a positive integer too)
y² + 88 = x²
x² - y² = 88
(x-y)(x+y) = 88

Factor 88:   1, 88 or 2, 44 or 4, 22 or 8, 11 

 

 

substitute n + 10 = x
(n + 10 - y)(n+ 10 +y) = 88

 

consider the cases

case 1
n + 10 - y = 1
n+ 10 + y = 88
add both of them equations, 2n + 20 = 89 ⇒ n = 34.5 this is not a solution because n is an integer

case 2
n + 10 - y = 2
n+ 10 +y = 44
add both of them equations, 2n + 20 = 46 ⇒ n = 13 this is solution because n is an integer


case 3
n + 10 - y = 4
n+ 10 +y = 22
add both of them equations, 2n + 20 = 26 ⇒ n = 3 this is solution because n is an integer

case 4
n + 10 - y = 8
n+ 10 +y = 11
add both of them equations, 2n + 20 = 19 ⇒ n = -0.5 this is not solution because n is an positive integer

So, the only possible values are 3 and 13 so the sum total of all possible n's is 16