Why is every answer of $5^k - 2^k$ divisible by 3?

Question

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 $5-2=3$?

Answer 1

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.

Answer 2

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.

Answer 3

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 $(\star)$.

Answer 4

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

$$5^k - 2^k \equiv 0 \pmod 3.$$

Answer 5

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}

Answer 6

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 $$5^k-2^k= {k \choose 1}2^{k-1}3 +\cdots +{k\choose 1}3^{k-1}2+ {k\choose k}3^k= 3\left ({k \choose 1}2^{k-1} +\cdots +{k\choose 1}3^{k-2}2+ {k\choose k}3^{k-1}\right)$$

Answer 7

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} ) $