How can I prove $1^{n+2} + 2^{n+2} + 3^{n+2} + 4^{n+2}$ is divisible by $10$ for any odd $n$?

Question

Assuming this is true:

$1^n+2^n+3^n+4^n$ divisible by $10$ for any odd $n$ ($n$ is natural)

How can I prove that for $n+2$:

$1^{n+2} + 2^{n+2} + 3^{n+2} + 4^{n+2}$

Is divisible by 10 as well ?

Thanks in advance

Answer 1

Induction:
Consider the difference, $$\left(1 + 2^{n+2} + 3^{n+2} + 4^{n+2}\right)-\left(1+2^n+3^n+4^n\right).$$ Is it divisible by $10$?

Answer 2

Another approach is to use congruences: First note that if $k$ is odd, then $$ 4^k\equiv 4\ (\text{mod}\,10). $$ Now, if $n$ is odd, then $n+2$ is odd, so $$ 1^{n+2}+2^{n+2}+3^{n+2}+4^{n+2}\equiv 1+2^{n+2}+3^{n+2}+4=5+2^{n+2}+3^{n+2}\ (\text{mod}\,10). $$ It suffices to prove that $5\mid 2^{n+2}+3^{n+2}$. Indeed, since $n+2$ is odd we get $$ 2^{n+2}+3^{n+2}\equiv 2^{n+2}+(-2)^{n+2}= 2^{n+2}-2^{n+2}=0\ (\text{mod}\,5). $$ Therefore $$ 5\mid 1^{n+2}+2^{n+2}+3^{n+2}+4^{n+2}. $$ Since clearly $$ 2\mid 1^{n+2}+2^{n+2}+3^{n+2}+4^{n+2} $$ the assertion follows.