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KnifeMissile · 02-09-2004, 12:26 PM

Okay, fine. Since so many people were so interested in this problem, I'll just post the solution right here!

You want to add the identity, 0, but in a very convenient form.

Let S be the summation symbol, represented in this clumsy ASCII form like so:

S(i=0,k)(a^i) = a^0 + a^1 + ... + a^k

So, add the identity 0 = S(i=1,n-1)(a^i) - S(i=1,n-1)(a^i) and the formula will simplify right before your eyes!

Code:

a^n - 1 = a^n - 1 + 0
        = a^n - 1 + S(i=1,n-1)(a^i) - S(i=1,n-1)(a^i)
        = S(i=1,n-1)(a^i) + a^n - 1 - S(i=1,n-1)(a^i)
        = a( S(i=0,n-2)(a^i) + a^(n-1)) - S(i=0,n-1)(a^i)
        = a*S(i=0,n-1)(a^i) - S(i=0,n-1)(a^i)
        = ( S(i=0,n-1)(a^i) ) * (a - 1)

As you can see, a^n -1 can be decomposed into two factors, S(i=0,n-1)(a^i) and a - 1. But a - 1 is supposed to be prime, so the only way that these two terms can not be factors is if a - 1 = 1.

Therefore, a = 2.
QED.

Pretty cool, eh!

02-09-2004, 12:26 PM	#7 (permalink)
KnifeMissile Location: Waterloo, Ontario	Okay, fine. Since so many people were so interested in this problem, I'll just post the solution right here! You want to add the identity, 0, but in a very convenient form. Let S be the summation symbol, represented in this clumsy ASCII form like so: S(i=0,k)(a^i) = a^0 + a^1 + ... + a^k So, add the identity 0 = S(i=1,n-1)(a^i) - S(i=1,n-1)(a^i) and the formula will simplify right before your eyes! Code: a^n - 1 = a^n - 1 + 0 = a^n - 1 + S(i=1,n-1)(a^i) - S(i=1,n-1)(a^i) = S(i=1,n-1)(a^i) + a^n - 1 - S(i=1,n-1)(a^i) = a( S(i=0,n-2)(a^i) + a^(n-1)) - S(i=0,n-1)(a^i) = aS(i=0,n-1)(a^i) - S(i=0,n-1)(a^i) = ( S(i=0,n-1)(a^i) ) (a - 1) As you can see, a^n -1 can be decomposed into two factors, S(i=0,n-1)(a^i) and a - 1. But a - 1 is supposed to be prime, so the only way that these two terms can not be factors is if a - 1 = 1. Therefore, a = 2. QED. Pretty cool, eh!