Sum of digits (upto single digit) of a prime number raised to an even power

Question

I noticed that sum of digits (upto a single digit) of a prime number (except 3) raised to an even power is always either 1, 4 or 7 (3 is an exception. 3 raised to any power always gives the sum of digits upto single digit as 9).

The primes which have sum of their digits upto single digit as 1 or 8 always give sum of digits upto single digit (when raised to even power) as 1. Those primes which give 4,5 or 7 as the sum of digits upto single digit (when squared) follow this pattern of 4,7,1 (when sum is 7) and 7,4,1 (when sum is 4 or 5) i.e, after every 3rd even power the pattern repeats(given in example below).

Has this been observed before? Is this useful somehow? Can anyone prove or disprove this?

Examples

1. PRIMES WITH THEIR SUM OF DIGITS UPTO SINGLE DIGIT AS 1 OR 8

   $19^2 = 361 => 3+6+1 = 10 => 1+0 = 1$

   $19^{24} = 48,98,76,29,30,96,08,46,81,77,16,29,52,77,921 => $ $4+8+9+8+...+9+2+1 = 163 => 1+6+3 = 10 => 1+0=1$

   $233^4 = 2,94,72,95,521 => 2+9+4+7+2+9+5+5+2+1 = 46 => $ $4+6 = 10 => 1+0 = 1$

   $233^{10} = 4,71,58,41,61,16,44,22,54,29,70,449 => $ $4+7+1+5+...+4+4+9 = 100 => 1+0+0 = 1$

2. PRIMES WITH THEIR SUM OF DIGITS UPTO SINGLE DIGIT AS 4 OR 5 OR 7 (Here I am giving only 13 as an example but it works for primes with sum of their digits upto single digit as 5 or 7 as well)

   $13^2 = 169 => 1+6+9 = 16 => 1+6 = 7 = 3+4 <= 34 = $ $8+1+5+...+2+1 <= 815730721 = 13^8$

   $13^4 = 28561 => 2+8+5+6+1 = 22 => 2+2 = 4 = 1+3 <= 13 $ $= 6+7 <= 67 = 1+3+7+...+8+4+9 $ $<= 1,37,85,84,91,849 = 13^{10}$

   $13^6= 4826809 => 4+8+2+6+8+0+9 = 37 => 3+7 = 10 =>$ $ 1+0 = 1 =1+0 <= 10 = 5+5 <= 55 $ $= 2+3+...+1 <=2,32,98,08,51,22,481 = 13^{12}$

Answer 1

Your procedure of calculating a sum of digits of $n$ over and over, until one digit is left, gives just the remainder of dividing $n$ by $9$ (where $0$ is substituted by $9$). This fact follows from a well-known divisibility test: the remainder of a number after dividing by $9$ is equal to the reminder of the sum of its digits. On each step, the result gets smaller, while maintaining its value modulo $9$. So it will become single-digit eventually. All single-digit natural numbers have different remainders modulo $9$.

Now, if we raise an integer to the second power, what remainders modulo $9$ can we get? $$0^2=0,$$ $$1^2=(-1)^2=1,$$ $$2^2=(-2)^2=4,$$ $$3^2=(-3)^2=0,$$ $$4^2=(-4)^2=7.$$

It means that if you take $n^2$ for any integer $n$, not divisible by $3$, then it will give $1$, $4$ or $7$ after your procedure. $3$ raised to any power greater than $1$ will give $9$.