2,3,4,5 can form () 7, for coprime

2,3,4,5 can form () 7, for coprime

three

Given n & gt; 1, a & gt; 1, and an-1 is a prime number, find the value of a, and explain that n is also the solution process of a prime number

The common factor of a prime number is 1 and itself
When n = 2, a ^ n-1 = a ^ 2-1 = (A-1) (a + 1), as long as one is 1,
Then a ^ 2-1 is prime, and ∵ a > 0, so obviously a + 1 ≠ 1, so A-1 = 1, a = 2
② When n = 3, a ^ n-1 = a ^ 3-1 = (A-1) (a ^ 2 + A + 1), similarly, ∵ a > 0
The latter factor ≠ 1, that is, a = 2
③ When n = 4, a ^ n-1 = a ^ 4-1 = (A-1) (a + 1) (a ^ 2 + 1), it is obviously a composite number,
∵ there are three factors, so we can't find a
④ When n = 5, a ^ n-1 = a ^ 5-1 = (A-1) (a ^ 4 + A ^ 3 + A ^ 2 + A + 1), the same as ②,
a-1=1,a=2
⑤ When n = 6, a ^ n-1 = a ^ 6-1 = (A-1) (a ^ 2 + A + 1) (a ^ 3 + 1), the same as ③ and ④
We can't find a
⑥ When n = 7, a ^ n-1 = a ^ 7-1 = (A-1) (a ^ 6 + A ^ 5 + A ^ 4 + A ^ 3 + A ^ 2 + A + 1),
The same as ②, ③, ⑤, A-1 = 1, a = 2
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To sum up, we can find a rule, that is, the number of a (a > 1)
If n (n > 1) is a composite number, then n can be expressed as n = XY (x, y are not 1)
Then a ^ n can be expressed as a ^ XY, that is, (a ^ x) ^ y, then a ^ x must be in the above law
So if n is a composite number, then a ^ n - 1 must be a composite number
In addition, we can find a rule that the fixed value of a is always 2, otherwise
If A-1 must be greater than 1, then a is greater than 2. If a is greater than 2, then the original number is a composite number
The prime factor is not 1 and itself
So here we come to the conclusion: a (a greater than 1) = 2, n is prime

An is the sum of the first n prime numbers. It is proved that there is at least one complete square number in [an, an + 1]

Let the nth prime number be p (n). Obviously, when n ≥ 2, P (n + 1) ≥ P (n) + 2. When n = 1, the original proposition holds. When n ≥ 2, let the largest complete square less than a (n) be K & # 178;, then (K + 1) &# 178; ≥ a (n). To prove the original proposition, we only need to prove (K + 1) &# 178; ≤ a (n + 1) and a (n + 1) = a (n) + P (n + 1), (k + 1) &# 178