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#1 2007-03-11 13:51:48

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

A problem

For

consider the rational number

Where

and   
are positive integers. Prove that if p is an odd prime , then the numberator
of
is divisible by p.


Numbers are the essence of the Universe

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#2 2007-03-11 14:00:13

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: A problem

I thought


and by wilson's theorem
if p is a prime
then get 

then I don't know what to do

Last edited by Stanley_Marsh (2007-03-11 14:02:39)


Numbers are the essence of the Universe

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#3 2007-03-11 14:56:09

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: A problem

You can’t do it that way – h[sub]p−1[/sub] is not an integer. shame

Last edited by JaneFairfax (2007-03-11 15:29:42)

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#4 2007-03-11 16:30:10

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: A problem

It’s actually ­really simple. Instead of taking an odd prime p, we’ll take any odd number n = 2a+1 > 1.

Now we rewrite the terms in the second half-sum backwards. Note that 2a = n−1, 2a−1 = n−2 … a+1 = na.

Note that for all k = 1, 2 … a

Therefore

for k = 1, 2 … a. This shows that u[sub]n−1[/sub] is divisible by n (so the result is true for all odd integers greater than 1, not just odd primes).

up

Last edited by JaneFairfax (2007-03-11 17:22:15)

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