Proof: if n > 1 and a ^ n - 1 is prime, then a = 2 and N is prime

Proof: if n > 1 and a ^ n - 1 is prime, then a = 2 and N is prime

A is a positive integer
a^n-1
=(a-1)[a^(n-1)+…… +a+1]
If a > = 3, A-1 > = 2
Now there is a factor A-1, not a prime
So only when a = 2 can it be prime
If n is not prime, n = PQ,
A ^ n-1 is divisible by (a ^ p-1) and (a ^ Q-1), not prime
So n is prime

How to prove that a Λ n-1 is prime, then a = 2 and N = P (P denotes prime)

First, decompose the factor
a^n-1=(a-1)(a^(n-1)+…… +1)
If a ^ n-1 is a prime, then A-1 = 1, so a = 2;
If n is not a prime, let n = st, s > 1, t > 1, then
a^n-1=(a^s-1)(a^(s(t-1))+…… +1) It's a combination, it's a contradiction
So n is prime
Proof of proposition

On the guess of prime number, help me prove it
Conjecture: suppose a is a natural number greater than 3, then there must be B to make a + B and A-B prime respectively
Note: the origin of this thing
I had a conjecture before - let 2m be an even number greater than 6, then there must be two prime numbers in the range of M + 3 and M-3, and the sum of them is 2m
This new conjecture is an enhanced version of conjecture, which expert can help me prove it!

Your conjecture is equivalent to Goldbach's conjecture. The original form of Goldbach's conjecture: any even number greater than 6 can be expressed as the sum of two prime numbers. Suppose Goldbach's conjecture holds. Let a be a natural number greater than 3, then 2a is an even number greater than 6, so there exists a prime number P, Q such that 2A = P + Q. let B = P-A, then a + B = P, A-B = a - (P-A)