Proving $3^n + 4^n + 5^n$ is divisible by $12$ for all positive odd integers $n$ by induction I am trying to prove by induction that $\sum_{i=3}^{5} i^n$ is divisible by $12$ for all positive odd integers $n$. I heard that the solution might involve normal induction, where you assume that it works for the lowest positive odd number (in this case, $1$) and then try to deduce $P(n+2)$ for the next odd number.
I am not sure how to go about this, so any help would be greatly appreciated. I was thinking that if we define $n$ to be odd, then we could try to show that $n+2$ works using the fact that $n+2$ is also odd. However, I am not sure if this is the correct approach or how to proceed.
Could someone please help me with this problem?
Thank you!

Proof 1 with Induction :
We know $X=3^{1}+4^{1}+5^{1}=12$ is Divisible by $12=3 \times 4$.
Let $Y=3^{n}+4^{n}+5^{n}$ be Divisible by $12$.
Let $Z=3^{n+2}+4^{n+2}+5^{n+2}$ , where we want to check Divisibility by $12$ , starting with $3$.
$Z=3^{2} \times 3^{n}+4^{2} \times 4^{n}+5^{2} \times 5^{n}$
$Z=9 \times 3^{n}+16 \times 4^{n}+25 \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n} = 3^{n}+4^{n}+5^{n}+3c=Y+3c$ , which must be Divisible by $3$
Continuing with $Z=3^{n+2}+4^{n+2}+5^{n+2}$ , where we want to check Divisibility by $12$ , which contains $4$.
$Z=3^{2} \times 3^{n}+4^{2} \times 4^{n}+5^{2} \times 5^{n}$
$Z=9 \times 3^{n}+16 \times 4^{n}+25 \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n} = 3^{n}+4^{n}+5^{n}+4k=Y+4k$ , which must be Divisible by $4$
Hence overall, $Z$ must be Divisible by $3 \times 4 = 12$.
With $X$ , $Y$ & $Z$ being Divisible by $12$ , we can see that the Induction Proof works here.
More generally , we can see that $(W-1,W,W+1)$ with $W$ a multiple of $4$ will have that Property.
With $W=4$ we get $(3,4,5)$ here.
With $W=8$ we get $(7,8,9)$ which will have that Property.
Proof 2 without Induction :
$(W-1)^n+(W)^n+(W+1)^n$ has 3 consecutive numbers where we must have a multiple of $3$ , the others must have "1 less than a multiple of $3$" & "1 more than a multiple of $3$".
Evaluating each Power will give terms with multiples of $3$ except the last terms which will be $(-1)^{ODD}$ & $(+1)^{ODD}$ which will give $-1+1=0$ hence the whole must be Divisible by $3$.
$(W-1)^n+(W)^n+(W+1)^n$ with $W=4w$ gives :
$(4w-1)^n+(4w)^n+(4w+1)^n$ with $W=4w$
When we evaluate each Power , we will get multiples of $4$ throughout , except for the last terms which will be $(-1)^{ODD}$ & $(+1)^{ODD}$ which will give $-1+1=0$ , hence the whole will be Divisible by 4.
In general , the whole must be Divisible by $3 \times 4 = 12$
In Particular , $W=4$ gives $3^n+4^4+5^n$ Divisible by $12$ when $n$ is ODD.
