For how many positive integers N is ⌊N^2/5⌋(floor of N^2/5) a prime?
To find the number of positive integers N for which ⌊N^2/5⌋ is a prime, we need to consider the possible values of N.
⌊N^2/5⌋ is the largest integer that is less than or equal to N^2/5. So, if ⌊N^2/5⌋ is a prime, it means that N^2/5 is not an integer and has only two integer divisors: 1 and ⌊N^2/5⌋ itself.
Let's analyze the possible values of N:
1. If ⌊N^2/5⌋ = 2:
This implies that N^2/5 is between 2 and 3 (exclusive). Hence, the possibilities for N are:
N^2/5 = 2.1 => N^2 = 10.5 => No integer solutions
N^2/5 = 2.2 => N^2 = 11 => No integer solutions
N^2/5 = 2.3 => N^2 = 11.5 => No integer solutions
Therefore, ⌊N^2/5⌋ cannot be 2.
2. If ⌊N^2/5⌋ = 3:
This implies that N^2/5 is between 3 and 4 (exclusive). Hence, the possibilities for N are:
N^2/5 = 3.1 => N^2 = 15 => No integer solutions
N^2/5 = 3.2 => N^2 = 16 => N = 4 or -4
N^2/5 = 3.3 => N^2 = 15.75 => No integer solutions
Therefore, ⌊N^2/5⌋ can be 3 only when N = 4 or -4.
3. If ⌊N^2/5⌋ = 5:
This implies that N^2/5 is between 5 and 6 (exclusive). Hence, the possibilities for N are:
N^2/5 = 5.1 => N^2 = 25 => N = 5 or -5
N^2/5 = 5.2 => N^2 = 26 => No integer solutions
N^2/5 = 5.3 => N^2 = 26.5 => No integer solutions
Therefore, ⌊N^2/5⌋ can be 5 only when N = 5 or -5.
From the analysis above, we can see that there are three positive integers N (4, 5, -5) for which ⌊N^2/5⌋ is a prime.
Hence, the answer is 3.