We know A is not 1. This can easily be seen as we are summing two positive values, p^{2} and 1007. Both of these values are already greater than 1. The smallest prime is 2, and 2 > 1, as is its square, 2^{2} = 4 > 1.

As every prime is an odd number, but 2, we know that p^{2} + 1007 is even and not 2, therefore A, itself, can not be prime.

It is known that the difference between any perfect square and its predecessor follows the identity, n^{2} = (n - 1)^{2} + (2n - 1). If A were to be a perfect square, given the identity, we can determine that n would equal 504 and p would be 503 (which does coincidentally happen to be prime). 2n - 1 = 1007 ==> n = 504 But as 504 is not prime, A can not be a perfect square as a solution.

As the product of two odd numbers is odd and the sum of two odd numbers is even, if A is a composite of two distinct primes and p is not 2, one of the composite primes would need be 2. Designating the other factor as b, we have A = 2b = p^{2} + 1007. From a quick check computing the values for b such that A is a positive integer, it is found that c is always even and greater than 2. This means b is not a prime, and thus A can not be a composite of two primes where one of the factors is 2. We are then left with 2 as a possibility for p: A = (2)^{2} + 1007 = 1011. The prime factors for 1011 are 3 and 337, giving a solution to our quest for four or less positive divisors (1, 3, 337, 1011) for A.