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   for prime >3, sum_k=1^(p-1) 1 over k =0 mod p^2
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   Author  Topic: for prime >3, sum_k=1^(p-1) 1 over k =0 mod p^2  (Read 1000 times)
4butipul2maro
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for prime >3, sum_k=1^(p-1) 1 over k =0 mod p^2  
« on: Sep 18th, 2008, 1:27am »
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Hello,
 
I heard that
for prime >3,  
sum_k=1^(p-1) 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^(p-1) 1 over k =0 mod  
« Reply #1 on: Sep 18th, 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/(p-1)=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 (p-1)!+(p-1)!/2+(p-1)!/3+...+(p-1)!/(p-1) (which will be true exactly if the result is true).
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Re: for prime >3, sum_k=1^(p-1) 1 over k =0 mod  
« Reply #2 on: Sep 18th, 2008, 2:33pm »
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This was discussed way back when.  But here is another argument:
pair off 1/k and 1/(p-k).
Divide through by p and work in the field Z/pZ.  Thus reduce to showing
1 + 1/22 + ... + 1/(p-1)2 = 0 mod p
.
Use the bijection k <-> 1/k on (Z/p)*.
« Last Edit: Sep 18th, 2008, 2:35pm by Eigenray » IP Logged
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