Prove (by induction) For n $\in \mathbf{N}\ 13^{n+1} − 7^n$ is divisible by 6

Question

I started with:
$n=0: 13^{0+1} - 7^0 = 13-1 = 12$ divisible by 6.
$n=1: 13^{1+1} - 7^1 = 162 $ divisible by 6.
$13^{n+1} − 7^n = 6*k$ for any $k \in \mathbf{N}$
$n \longrightarrow n+1:$
\begin{align*} 13^{(n+1)+1} − 7^{n+1} &= 13^{n+2} − 7^{n+1}\\ &= 13^{n+2} − 7^{n+1} \end{align*}

However this is where i'm stuck.

See here in the dupe for how to derive such inductive proofs in an intuitive arithmetical way - making the proof very simple. — Bill Dubuque, May 08 '22 at 19:47

score 2 · Accepted Answer · answered May 08 '22 at 17:44

2

Hint: You may write down $13=6+7$ and use the fact that $6\cdot k$ is divisible by $6$ for any $k\in \mathbb{Z}$.

The solution is as follows, $13^{n+2}-7^{n+1}=13\cdot 13^{n+1}-7\cdot 7^n=(6+7)\cdot 13^{n+1}-7\cdot 7^n=6\cdot 13^{n+1}+7(13^{n+1}-7^n)$. The last expression is divisible by $6$ by the induction hypothesis.

answered May 08 '22 at 17:44

TT_16

81

1

See here for the arithmetical intuition behind such inductive proofs. – Bill Dubuque May 08 '22 at 19:46

score 0 · Answer 2 · answered May 08 '22 at 17:58

0

Since $13=1 \mod 6$ and $7=1 \mod 6 $ we have that $13^n-7^n=0 \mod 6$

answered May 08 '22 at 17:58

Leox

8,120

Prove (by induction) For n $\in \mathbf{N}\ 13^{n+1} − 7^n$ is divisible by 6

2 Answers2