For every positive integer $n$, let $D_n$ be the greatest common divisor all numbers of form $a^n+(a+1)^n+(a+2)^n$,$a$ ranges between all positive integers. Prove that, for every $n$, $D_n$ is of the form $3^k$ for some integer $k\ge0$.
How can solve this problem without using the induction principle?
we have $D_n$ divides all elements of the form $a^n+(a+1)^n+(a+2)^n$ so for any $a$ we have : $$ D_n \text{ divides }a^n+(a+1)^n+(a+2)^n\\ D_n \text{ divides }(a+1)^n+(a+2)^n+(a+3)^n$$
so for any $a$ we have $D_n$ divides $(a+3)^n-a^n$ and now set $a=D_n$ and conclude that $D_n$ divides $(D_n+3)^n$ but $(D_n+3)^n\equiv 3^n \mod D_n$
When you read
$D_n$ is of the form $3^k$ for some integer $k≥0$
you should actually read
$D_n$ can't be a multiple of any prime $p\neq 3$
Now consider separately the case $p=2$ and $p>3$ and you are done:
For 2, just take $a$ even. Then $a^n$ is even, $(a+1)^n$ is odd and $(a+2)^n$ is even, so their sum is odd.
For $p>2$ and $n$ even, take $a=p-1$. Working mod $p$, $a^n\equiv 0$, $(a+1)^n\equiv 1$ and $(a-1)^n \equiv 1$ and the sum is $2\not\equiv 0$.
For $p>3$ and $n$ odd, take $a=p$ suppose $p$ divides $p^n+(p+1)^n+(p+2)^n \equiv 1 + 2^n$. Now take $a=p+1$, and you'll get that $p$ divides $1+2^n+3^n$ but that's impossible as $F_p$ is a field and $3^n\not\equiv 0$ (except for the case $p=3$)
Write $\ f(a) = (a\!+\!2)^n+(a\!+\!1)^n+a^n.\ $ By basic Congruence Rules $ $
$\begin{array}{} {\rm mod}\ D\!:\!\! &f(1)\equiv f(1\!+\!D)\equiv 0\\ &f(0)\equiv f(0\!+\!D)\equiv 0\end{array}\Rightarrow\, 0\equiv f(1)\!-\!f(0)\equiv 3^n\!+2^n\!+1^n\!-(2^n\!+\!1^n\!+0^n)\equiv \color{#c00}{3^n}$
Therefore $\,D = D_n\mid\color{#c00}{ 3^n}$
