It is known that for any natural number m, N, f (m) f (n) = f (n + m) + F (n-m) It is known that for any natural number m, N, there is f (m) f (n) = f (n + m) + F (n-m), where f (0) ≠ 0, f (1) = 1, find f (n) ———————————————————————————————————————— -------------------------- Sorry, cos n π

It is known that for any natural number m, N, f (m) f (n) = f (n + m) + F (n-m) It is known that for any natural number m, N, there is f (m) f (n) = f (n + m) + F (n-m), where f (0) ≠ 0, f (1) = 1, find f (n) ———————————————————————————————————————— -------------------------- Sorry, cos n π

This topic itself is more interesting: another M = 1, f (n) = f (n + 1) + F (n-1) if the trigonometric function is better, we may see how to construct it. (I constructed it when I guessed the answer) there are: 2cos [(a + b) / 2] * cos [(a-b) / 2] = cos a - cos B, we can see that compared with the known conditions, 2 * cos [(a-b) / 2] = cos a - cos B can be seen

If the natural number m, N, m x power + N Y power = 17 Z power and 1 / x + 1 / y = 1 / Z, then M + n=
`

5
1^2+4^2=17^1
m=1；n=4；x=2；y=2；z=1；

Let n square matrix, a not equal to 0, m power of a equal to 0, find the eigenvalue of A
It is proved that a is not similar to diagonal matrix

Let a be the eigenvalue of a, then for any polynomial f, if f (a) = 0, then f (a) = 0 (eigenvalues are the roots of the smallest polynomial, and the smallest polynomial divides the arbitrarily zeroed polynomial, so the eigenvalue is the root of the arbitrarily zeroed polynomial). Now f (a) = a ^ m = 0, so f (a) = a ^ m = 0, there must be a = 0