That's a tricky one
Number Theory
Level
4
The number of positive integral values of for which is a prime number is
2
1
4
3
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1 solution
Factorizing we get ,
f ( n ) = n 3 − 8 n 2 + 2 0 n − 1 3 = ( n − 1 ) ( n 2 − 7 n + 1 3 )
Since it is a prime any one of its factors must be 1.
Case 1: n − 1 = 1 ⟹ n = 2 Case 2: n 2 − 7 n + 1 3 = 1 ⟹ ( n − 4 ) ( n − 3 ) = 0 ⟹ n = 4 , 3
We observe that f ( 2 ) , f ( 3 ) , f ( 4 ) are primes thus there are 3 values.