Question

For how many positive integers $n$, $n^3-8n^2+20n-13$ is a prime number?

Answer 1

$n^3-8n^2+20n-13$, where $n$ is a positive integer.
For, $n=1$, we get
$=1^3-8\times (1{)}^2+20\times 1-13$
$=1-8+20-13$
$=21-21$
$=0$ (Not a prime number)
For, $n=2$, we get
$=2^3-8\times (2{)}^2+20\times 2-13$
$=8-32+40-13$
$=48-45$
$=3$ ( A prime number)
For, $n=3$, we get
$=3^3-8\times (3{)}^2+20\times 3-13$
$=27-72+60-13$
$=87-85$
$=2$ ( A prime number)
For, $n=4$, we get
$=4^3-8\times (4{)}^2+20\times 4-13$
$=64-128+80-13$
$=144-141$
$=3$ ( A prime number)
For, $n=5$, we get
$=5^3-8\times (5{)}^2+20\times 5-13$
$=125-200+100-13$
$=225-213$
$=12$ ( Not a prime number, but an even number)
For, $n=6$, we get
$=6^3-8\times (6{)}^2+20\times 6-13$
$=216-288+120-13$
$=336-301$
$=35$ ( Not a prime number, but an odd number)
So, we can say that there are only $3$ positive integers which give us prime numbers