a. B and N are fixed natural numbers, and for any natural number k (K ≠ b), A-K ^ n can be divisible by B-K, it is proved that a = B ^ n

a. B and N are fixed natural numbers, and for any natural number k (K ≠ b), A-K ^ n can be divisible by B-K, it is proved that a = B ^ n

Let f (k) = A-K ^ n
∵ f (k) contains the factor (B-K). According to the remainder theorem, f (b) = 0, ∵ A-B ^ n = 0, ∵ a = B ^ n

M n are all non-zero natural numbers. If M is greater than N, we arrange Mn, M2 and N2 from large to small

We can assume that M = 3, n = 2
mn=6
m²=9
n²=4
m²>mn>n²

It is known that M2 = n + 2, N2 = m + 2 (m ≠ n). Find the value of M2 + 2Mn + N2

By subtracting the known two formulas, it is obtained that m2-n2 = N-M, ∵ m ≠ n, ∵ m + n = - 1, ∵ M2 + 2Mn + N2 = (M + n) 2 = (- 1) 2 = 1;