Guest Posted October 17, 2009 Report Share Posted October 17, 2009 Determine all possible prime(s) P such that the total number of positive divisors of P2 + 1007, including 1 and itself, is less than 7. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 17, 2009 Report Share Posted October 17, 2009 P=2. This is the only solution... Let Q = P2 + 1007 If P is odd, P2 is odd and can be written in the form 8n+1. 1007 can be written as 8m+7 and so Q = 8m + 8n + 8. Hence, Q will have factors 1, 2, 4, 8, Q/8, Q/4, Q/2, Q. This is a minimum of 8 factors. Therefore there are no other solutions. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 18, 2009 Report Share Posted October 18, 2009 (edited) P=2. This is the only solution... Let Q = P2 + 1007 If P is odd, P2 is odd and can be written in the form 8n+1. 1007 can be written as 8m+7 and so Q = 8m + 8n + 8. Hence, Q will have factors 1, 2, 4, 8, Q/8, Q/4, Q/2, Q. This is a minimum of 8 factors. Therefore there are no other solutions. Well done, PM. Your proof is almost identical to mine, except that I started out with assuming P as a prime number throughout. It is relatively easy to prove that P2 1 is divisible by 24, whenever P is a prime number >=5. Since 1008 is a multiple of 24, it follows that: P2 1 + 1008 = P2 + 1007 is a multiple of 24 whenever P is a prime >= 5. Since 24 has precisely 8 factors, which is >7, this contravenes the given conditions. It is now a simple matter to substitute P = 2, 3 in the expression P2 + 1007. Checking the number of divisors in each, only P=2 satisfies the criterion that total number of positive divisors of P2 + 1007 is less than 7. Therefore, P=2 is the only possible solution. However in my opinion, your proof is better than mine, being shorter and concise. Edited October 18, 2009 by K Sengupta Quote Link to comment Share on other sites More sharing options...
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Determine all possible prime(s) P such that the total number of positive divisors of P2 + 1007, including 1 and itself, is less than 7.
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