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        riddles >> putnam exam (pure math) >> for prime >3, sum_k=1^(p-1) 1 over k =0 mod p^2
        
(Message started by: 4butipul2maro on Sep 18^th, 2008, 1:27am)

Title: for prime >3, sum_k=1^(p-1) 1 over k =0 mod p^2
Post by 4butipul2maro on Sep 18^th, 2008, 1:27am

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 ~

Title: Re: for prime >3, sum_k=1^(p-1) 1 over k =0 mod
Post by Obob on Sep 18^th, 2008, 11:21am

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).

Title: Re: for prime >3, sum_k=1^(p-1) 1 over k =0 mod
Post by Eigenray on Sep 18^th, 2008, 2:33pm

This was discussed [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1089736078]way back when[/link]. But here is another argument:
[hide]pair off 1/k and 1/(p-k)[/hide].
[hide]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[/hide].
[hide]Use the bijection k <-> 1/k on (Z/p)^*[/hide].