Question

For how many positive integers n. N^3-8*^2+20n-13 is a prime.

Answer 1

n^3- 8n^2+20n -13

= (n-1)(n^2 - 7n + 13 )

In order to make (n-1) prime it will either be +/-1 or (n^2 - 7n + 13 ) has to be +/-1

n-1 = +/-1

= n = 0 or 2

n^3- 8n^2+20n -13 = -13 or 3

(n^2 - 7n + 13 ) = +/-1

= n is 3 or 4.

For this we will get n^3- 8n^2+20n -13 = 2 or 3

We know that prime numbers are positive so the prime numbers will be 2,3 and 4.

Answer 2

Answer:

n^3- 8n^2+20n -13

= (n-1)(n^2 - 7n + 13 )

If it has to be a prime, (n-1) has to be either +/-1 or (n^2 - 7n + 13 ) has to be +/-1

Case 1 : n-1 = +/-1

=> n = 0 or 2.

In that case, n^3- 8n^2+20n -13 = -13 or 3

Case 2: (n^2 - 7n + 13 ) = +/-1

=> n is 3 or 4.

In that case, n^3- 8n^2+20n -13 = 2 or 3

Considering that prime numbers are +ve, I would say that there are 3 values of n which make the expression prime : 2, 3, 4