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FOUR OR LESS (Posted on 2010-07-05) Difficulty: 2 of 5
Determine all prime numbers p such that the total number of positive divisors of A = p^2 + 1007 (including 1 and A) is 4 or less.

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Solution an analytical solution (at least an attempt) | Comment 1 of 3
If a positive number has 4 divisors or less it need be
  1. 1
    [1 divisor: (1)]
  2. be itself prime
    [2 divisors: (1, p)]
  3. be a perfect square of a prime
    [3 divisors: (1, p, p2)]
         - or - 
  4. be a composite of two distinct primes
    [4 divisors: (1, a, b, ab)]
  1. We know A is not 1. This can easily be seen as we are summing two positive values, p2 and 1007. Both of these values are already greater than 1. The smallest prime is 2, and 2 > 1, as is its square, 22 = 4 > 1.
  2. As every prime is an odd number, but 2, we know that
    p2 + 1007 is even and not 2, therefore A, itself, can not be prime.
  3. It is known that the difference between any perfect square and its predecessor follows the identity,
    n2 = (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.
  4. 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 = p2 + 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.

    Thus, the only solution for p is 2.

  Posted by Dej Mar on 2010-07-05 14:02:16
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