Higher powers of 5

Each cell is one n, coloured by P(n) mod m (grey = 0). Zoom in with the slider, drag to move around, hover to read a value. The image download contains all the rows shown, at full resolution.

Measured

Proved

Period and mirror. For m = 5^j: P(n + 5^j+1) − P(n) = 2n·5^j + 5^2j+1, which is divisible by 5^j, so 5^j+1 is a period; and P(−n) = P(n), so each period is a mirror image of itself. (Minimality is checked by computer.)

Zeros: 5^⌈j/2⌉ + 4 per period. P(n) ≡ 0 (mod 5^j) means n² + 4n5 ≡ 0 (mod 5^j+1). For n = 5t this is 5t² ≡ 0 (mod 5^j), true for exactly 5^⌈j/2⌉ values of t in a period. Otherwise it reads n² ≡ 4 or n² ≡ −4 (mod 5^j+1); each has exactly two solutions (they lift uniquely from n ≡ ±2 and n ≡ ±1 mod 5), and those solutions automatically have the right sign of n5. Total: 5^⌈j/2⌉ + 4, giving 9, 9, 29, 29, 129, 129, …