It is proved by mathematical induction that a ^ (n + 1) + (a + 1) ^ (2n-1) can be divided by a ^ 2 + A + 1, and N belongs to positive integer

When n = 1, a ^ 2 + (a + 1) satisfies
When n = k, a ^ (K + 1) * a + (a + 1) ^ (2k-1) * a can be divisible by a ^ 2 + A + 1
When n = K + 1
a^(k+1+1) +(a+1)^(2k+2-1)
= a^(k+1) *a + (a+1)^(2k-1)(a+1)^2
= a^(k+1) *a +(a+1)^(2k-1) (a^2+2a+1)
=a^(k+1) *a+(a+1)^(2k-1) * a + (a+1)^(2k-1) (a^2+a+1)
Obviously, the left and right parts of the above formula can be divided by a ^ 2 + A + 1, so the whole formula can be divided by a ^ 2 + A + 1
Therefore, it is proved that

Try to explain: for any natural number n, the value of n (n + 5) - (n-3) (n + 2) can be divided by 6

∵ n (n + 5) - (n-3) (n + 2) = (N2 + 5N) - (n2-n-6) = N2 + 5n-n2 + N + 6 = 6N + 6 = 6 (n + 1) and N ≥ 1 ∵ can always be divisible by 6

There are four different natural numbers, among which the sum of any two numbers is a multiple of 2, and the sum of any three numbers is a multiple of 3. In order to make the sum of the four numbers as complete as possible

There are four different natural numbers, and the sum of any two of them is a multiple of 2. So the remainder of the four numbers divided by 2 is the same. So the remainder of the four numbers divided by 6 is the same, and the minimum is 1, 7, 13, 19. The sum is 40

There are four different positive integers, in which the sum of any two numbers is a multiple of 2, and the sum of any three numbers is a multiple of 3. To make the sum of the four numbers as small as possible, what are the four numbers?

The sum of any two numbers is a multiple of 2, which means that the four numbers are either multiples of 2 or not multiples of 2. The sum of any three numbers is a multiple of 3. This paper analyzes several hypotheses: 1. Suppose that the four numbers are multiples of 3 - the situation can be established; 2. Suppose that one of the four numbers is a multiple of 3 - the remaining three numbers are in pairs

It is proved by mathematical induction that a ^ (n + 1) + (a + 1) ^ (2n-1) can be divided by a ^ 2 + A + 1, and N belongs to positive integer