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| FOUR OR LESS (Posted on 2010-07-05) |
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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.
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Submitted by Ady TZIDON
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Rating: 4.0000 (1 votes)
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Solution:
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ans: P=2
SOLUTION:
Prime number p is either odd or 2.
Odd squared number equals 1 modulo 4, and 1007 is -1 mod 4, so A is divisible by 4.
This means it has at least 6 divisors: 1, 2, 4, A/4, A/2, A.
If p equals 2, 1011 (=4+1007=3*337), has exactly 4 factors, 1, 3, 337, 3*337.
Thus, p=2 is the only valid answer.
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