N is a natural number. How many values of N exist, such that N2+24N+21 has exactly three factors?
For N2+24N+21 to have exactly three factors, it must be square of a prime number.
Let N2+24N+21=a2 where a is a prime number.
⟹(N+12+a)(N+12-a) = 123
Either N+12+a = 123 and N+12-a = 1
Or N+12+a = 41 and N+12-a = 3
In the first case N = 50 and a = 61.
In the second case N =10 and a =19.
In either case N2+24N+21 is the square of a prime number.
So two such values exist.