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Topic: for prime >3, sum_k=1^(p1) 1 over k =0 mod p^2 (Read 1000 times) 

4butipul2maro
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for prime >3, sum_k=1^(p1) 1 over k =0 mod p^2
« on: Sep 18^{th}, 2008, 1:27am » 
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Hello, I heard that for prime >3, sum_k=1^(p1) 1 over k =0 mod p^2. In fact, it seens like true. For example, for p=5, 1/1 + 1/2 +1/3 +1/4 = 1 + 13 + 17 + 19 mod 25 = (1+19) + (13 +17) = 20 + 30 mod 25 = 4+6 mod 5 =0 mod 5. But generally i don's know. Help me ~


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Obob
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Re: for prime >3, sum_k=1^(p1) 1 over k =0 mod
« Reply #1 on: Sep 18^{th}, 2008, 11:21am » 
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The way you have stated this, it is confusing. It should be clarified that what you mean by 1/k is the multiplicative inverse of k in the ring Z/p^2Z. Then the identity you are trying to prove is simply 1+1/2+1/3+...+1/(p1)=0. And in your proof for p=5, you could write 1+13+17+19 = 50 = 0 (mod 25); being 0 mod 25 is more information than being 0 mod 5 is, so your argument doesn't technically conclude that the sum is 0 mod 25. The step "20 + 30 mod 25 = 4+6 mod 5" is also nonsense: 20+30 mod 25 after changing the modulus to 5 is still just 20+30 mod 5. I agree that the result seems true, from having tested many cases up to p=113. One natural thing to try is to show that p^2 divides the sum (p1)!+(p1)!/2+(p1)!/3+...+(p1)!/(p1) (which will be true exactly if the result is true).


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Eigenray
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Re: for prime >3, sum_k=1^(p1) 1 over k =0 mod
« Reply #2 on: Sep 18^{th}, 2008, 2:33pm » 
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This was discussed way back when. But here is another argument: pair off 1/k and 1/(pk). Divide through by p and work in the field Z/pZ. Thus reduce to showing 1 + 1/2^{2} + ... + 1/(p1)^{2} = 0 mod p. Use the bijection k <> 1/k on (Z/p)^{*}.

« Last Edit: Sep 18^{th}, 2008, 2:35pm by Eigenray » 
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