How I can prove that the last digit of $1+6^n+2\times 3^n+7^n+4^n+3\times9^{n}+4\times8^n$ is $3$ or $9$?

Question

I have checked the first $14$ digits of Golden ratio, and I have found some attractive properties. I have defined the sequence as

$6^n+1^n+8^n+0^n+3^n+3^n+9^n+8^n+8^n+7^n+4^n+9^n+8^n+9^n$.

Some properties I have noted are the following:

1/-The number: $61803398874989$ is a prime number, and the sum of digits of that number is also a prime number.

2/ $6^n+1^n+8^n+0^n+3^n+3^n+9^n+8^n+8^n+7^n+4^n+9^n+8^n+9^n$ is congruent $1$ or $3 \bmod 4$. It is a prime number for $n=1,2,37,61,70,81$ for $n<100$; for $n=3$ the term is cubic, and it can be written as $4913=17^3$.

3/The number $61^3+80^3+33^3+98^3+87^3+49^3+89^3$ is a prime number and it holds for $61^3+80^3+33^3+98^3+87^3+49^3-89^3$; for $n=57$ also it is a prime number.

4/The last digits of the sequence ($6^n+1^n+8^n+0^n+3^n+3^n+9^n+8^n+8^n+7^n+4^n+9^n+8^n+9^n$) is $3$ or $9$ using mathematica up to $n=10^6$ or just wolfram alpha.

Now I'm interested in this sequence, which is not mentioned in OEIS; I want to prove that really the last digit of $1+6^n+2\times 3^n+7^n+4^n+3\times9^{n}+4\times8^n$ is $3$ or $9$ .

Answer 1

The last digit is $3$, $9$, $3$, and $3$ for $n=1, 2, 3, $ and $4$, respectively.

Now can you prove $a^{n+4}\equiv a^n\bmod 10$ for $n\ge1$?

Hint: $2$ and $5$ divide $a^5-a$.

Answer 2

Taken mod $2$ we have

$$1+6^n+2\cdot3^n+7^n+4^n+3\cdot9^n+4\cdot8^n\equiv1+0+0+1+0+1+0\equiv1$$

Taken mod $5$ we have

$$1+6^n+2\cdot3^n+7^n+4^n+3\cdot9^n+4\cdot8^n\equiv1+1^n+2(-2)^n+2^n+(-1)^n+3(-1)^n+4(-2)^n\\ \equiv2+2^n+(-2)^n-(-1)^n\\ \equiv3+(2^n-1)(1+(-1)^n)\\ \equiv \begin{cases} 4&\text{if } n\equiv2\mod4\\ 3&\text{if } n\not\equiv2\mod4 \end{cases}$$

Putting these together, the last digit can only be $3$ or $9$, with the latter case occuring when $n\equiv2$ mod $4$ (i.e., for $n=2,6,10,14,\ldots$).