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Topic: Prime number problem (Read 1166 times) |
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Benny
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Prime number problem
« on: Jun 3rd, 2009, 12:42pm » |
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I'm not sure if this forum is the right one for this problem.
I have 2 prime numbers p and q and a natural number n that satisfy the following relationship
1/p + 1/q + 1/pq = 1/n
So, I rearrange the fractions
1/n = (p + q + 1)/pq
which gives me
n = pq /(p + q + 1)
n is a natural number, hence (p + q + 1) must divide pq.
I found this is true for
p = 2 and q = 3, or
p = 3 and q = 2
hence pq = 6
1/2 + 1/3 + 1/6 = 1, so n = 1
What about primes larger than 3.
How to prove that (p + q + 1) does not divide pq?
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SMQ
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Re: Prime number problem
« Reply #1 on: Jun 3rd, 2009, 1:19pm » |
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Ah, but you already know the prime factorization of pq, which is, trivially, p * q; so the only numbers which divide pq are 1, p, q, and pq. Clearly p + q + 1 is greater than any of 1, p, or q, so the only solution is p + q + 1 = pq. This implies p + 1 = pq - q = (p - 1)q --> q = (p + 1)/(p - 1) and since q is a prime, the only solutions are p = 2 or p = 3. Likewise p = (q + 1)/(q - 1) gives the same possibilities for q. Q.E.D.
--SMQ
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Benny
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Re: Prime number problem
« Reply #2 on: Jun 4th, 2009, 10:28am » |
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on Jun 3rd, 2009, 1:19pm, SMQ wrote:Ah, but you already know the prime factorization of pq, which is, trivially, p * q; so the only numbers which divide pq are 1, p, q, and pq. Clearly p + q + 1 is greater than any of 1, p, or q, so the only solution is p + q + 1 = pq. This implies p + 1 = pq - q = (p - 1)q --> q = (p + 1)/(p - 1) and since q is a prime, the only solutions are p = 2 or p = 3. Likewise p = (q + 1)/(q - 1) gives the same possibilities for q. Q.E.D.
--SMQ |
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Right. Thanks.
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