Show that if p is prime and p>3, then 2^p-2 +3^p-2 +6^p-2=1(mod p) I need help p
ID: 1943946 • Letter: S
Question
Show that if p is prime and p>3, then 2^p-2 +3^p-2 +6^p-2=1(mod p)I need help proving this with a proof. A proof solution gets the lifesaver rating..thanks
If you can explain the steps in this process thourouly it would be greatly appreciated.
Explanation / Answer
Assume that: 2^p-2 +3^p-2 +6^p-2 = x (mod p) we want to show that x = 1. Multiply the equation by 6: 6*(2^p-2 +3^p-2 +6^p-2 = x (mod p)) 3*2*2p-2 + 2*3*3p-2 + 6p-1 = 6x (mod p) 3*2p-1 + 2*3p-1 + 6p-1 = 6x (mod p) p>3 , thus: gcd(2,p) = 1 , gcd(3,p) = 1 so we have gcd(2*3,p) = gcd(6,p) = 1 Thus from the Fermat's little theorem we have: 2p-1 = 1 (mod p) 3p-1 = 1 (mod p) 6p-1 = 1 (mod p) Therefore: 3*2p-1 + 2*3p-1 + 6p-1 = 6x (mod p) 6x = 3*(1) + 2*(1) + 1 (mod p) 6x = 6 (mod p) p | (6x-6) p|6(x-1) and since gcd(6,p)=1 : p|(x-1) x = 1 (mod p) and we have: 2p-2 + 3p-2 + 6p-2 = 1 (mod p)
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