Why is every answer of $5^k - 2^k$ divisible by 3? We have the formula $$5^k - 2^k$$
I have noticed that every answer you get from this formula is divisible by 3. At least, I think so. Why is this? Does it have to do with $5-2=3$?
 A: Yes, it does!  
It's because in general you have the factorization: 
$$
x^k-y^k = (x-y)(x^{k-1}+x^{k-2}y+\dots+y^{k-1})
$$
Substituting in $x=5$ and $y=2$ should show you why that works.  
A: You say congruences are not familiar. Suppose you don't know the formulas in the other answers, but you do know the polynomial Division Algorithm. Dividing $\rm\:x^k\!-\!2^k$ by  $\rm\:x\!-\!2\:$ yields 
$\rm\qquad x^k-2^k =\ q(x)\, (x-2) + r\quad  $ for an integer $\rm\:r\:$ and integer coefficient quotient $\rm\:q(x).$
Evaluating at $\rm\ x = 2\ $ shows $\rm\ r = 0.\ $ Evaluating at $\rm\: x = 5\:$ shows $\rm\,3\,$ divides $\rm\,5^k - 2^k$ 
Remark $\ $ This is a special case of the Factor Theorem, as are the other answers.
A: Recall that $$a^k - b^k = (a-b)(a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1}) \tag{$\star$}$$
One way to see this is, to notice that
$$(a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1})$$ is the sum of first $k$ terms of a geometric progression with first term $a^{k-1}$ and common ratio $\dfrac{b}a$. We hence get that
$$a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1} = a^{k-1} \left(\dfrac{1-(b/a)^k}{1-b/a}\right) = \dfrac{a^k-b^k}{a-b}$$
Rearranging above gives you $(\star)$.
A: Even a little knowledge of modular arithmetic makes this obvious:
$$5 \equiv 2 \pmod 3$$
and thus
$$5^k \equiv 2^k \pmod 3,$$
which is equivalent to
$$5^k - 2^k \equiv 0 \pmod 3.$$
A: Let's use induction to prove that:
\begin{equation}
5^k - 2^k
\end{equation}
is divisible by 3
Let's check the proposition for k = 0 is divisible by 3
\begin{equation}
5^0 - 2^0 = 1 - 1 = 0
\end{equation}
which is divisible by 3
Let's assume that for k = n the proposition
\begin{equation}
5^n - 2^n
\end{equation}
is divisible by 3 and let's prove that
\begin{equation}
5^{(n+1)} - 2^{(n+1)}
\end{equation}
 is also true, so
\begin{equation}
    5^{(n+1)} - 2^{(n+1)} = 5{(5^n)} - 2{(2^n)}
\end{equation}
\begin{equation}
                      = (3 + 2)(5^n) - 2(2^n)
\end{equation}
\begin{equation}
                      = 3(5^n) + 2(5^n) - 2(2^n)
\end{equation}
\begin{equation}
                      = 3(5^n) + 2(5^n - 2^n)
\end{equation}
So in the above expression
\begin{equation}
    3(5^n)
\end{equation}
is divisible by 3
\begin{equation}
    2(5^n - 2^n)
\end{equation}
is divisible by 3 as per our assumption, so the finale expression
\begin{equation}
    3(5^n) + 2(5^n - 2^n)
\end{equation}
is also divisible by 3
So we can conclude that
\begin{equation}
    5^{(n+1)} - 2^{(n+1)}
\end{equation}
is divisible by 3
Meaning
\begin{equation}
    5^n - 2^n
\end{equation}
is divisible by 3 for
\begin{equation}
n \in \mathbb{N}
\end{equation}
A: Write $5^k= (2+3)^k= {k\choose 0}2^k+ {k \choose 1}2^{k-1}3 +\cdots +{k\choose 1}3^{k-1}2+ {k\choose k}3^k$ 
Therefore 
$$5^k-2^k= {k \choose 1}2^{k-1}3 +\cdots +{k\choose 1}3^{k-1}2+ {k\choose k}3^k=
3\left ({k \choose 1}2^{k-1} +\cdots +{k\choose 1}3^{k-2}2+ {k\choose k}3^{k-1}\right)$$
A: We know that $x^k −y^k$ is always divisible by $x-y$ due to formula $x^k −y^k =(x−y)(x^{k−1} +x^{k−2} y+⋯+y^{ k−1} ) $
