If 3 Λ n + M can be divisible by 13, try to explain that 3 Λ (M + 3) plus M can also be divisible by 13

If 3 Λ n + M can be divisible by 13, try to explain that 3 Λ (M + 3) plus M can also be divisible by 13

Is 3 ^ (n + 3) divisible by 13?
If so, look below
3^(n+3)+m=(3^n)*3^3+m=3^n*(26+1)+m=26*3^n+(3^n+m)
26 * 3 ^ n = 13 * (2 * 3 ^ n) can be divided by 13
(3 ^ n + m) can also be divisible by 13
So 3 ^ (n + 3) can be divisible by 13

It is proved that if 2 does not divide m and 3 does not divide m, then 24 does not divide m ^ 2 + 23
Change to 24 divisible m ^ 2 + 23

Let 2 not divide m, let m = 2A + 13 not divide m, let m = 3B + 1 or 3b-1m ^ 2 + 23 = (2a + 1) ^ 2 + 23 = 4A ^ 2 + 4A + 1 + 24 = 4A (a + 1) + 24a and a + 1 be adjacent natural numbers, so a (a + 1) can be divided by 2, then 4a (a + 1) can be divided by 8, 4a (a + 1) + 24 can be divided by 8, m ^ 2 + 23 = (3b + 1) ^ 2 + 23 = 9b ^ 2 + 6B + 1

It is proved by factorization that 257-512 can be divided by 120

It is proved that (512) = 1 × 512-257 = 1 × 512-257