I am looking for a proof of a certain set being divisible by 7

Question

I am looking for a proof of this statement:

$$7\mid{3^{6k+2}-{2^{6k+1}}}$$

By trial and error I can see that it holds but I cant figure out anyway to prove it or cant seem to be able understand why.

Any tips and help would be appreciated.

Answer 1

HINT: Write your term in the form $$(729)^k\cdot 9-(64)^k\cdot 2$$ and $$729\equiv 1\mod 7$$ and $$64\equiv 1 \mod 7$$ so we get $$729^k\cdot 9-64^k\cdot 2\equiv 1^k\cdot 9-1^k\cdot 2\equiv 9-2=7\equiv 0 \mod 7$$

Answer 2

Since, by Fermat's little theorem, $3^6\equiv1\pmod7$ and $2^6\equiv1\pmod7$, then$$3^{6k+2}-2^{6k+1}=(3^6)^k\times9-(2^6)^k-1\equiv2-1=1\pmod7.$$

Answer 3

$3^{6k+2}=(3^2)^{3k+1}=2^{3k+1}\mod 7$

$2^{6k+1}-2^{3k+1}=2^{3k+1}(2^{6k+1-3k-1}-1)=2^{3k+1}(2^{3k}-1)$

$2^{3k}-1=8^k-1=0\mod 7$

Answer 4

For $k=0$ we can actually verify that $7|3^2-2$.

Suppose that $3^{6k+2}-{2^{6k+1}}$ is divisible by $7$ for $k=n$. Then for $k=n+1$, $$ {3^{6k+2}-{2^{6k+1}}}=3^{6n+8}-{2^{6n+7}}=3^6(3^{6n+2}-2^{6n+1})+(3^6-2^6)\times2^{6n+1}\\=3^6(3^{6n+2}-2^{6n+1})+7\times 95\times2^{6n+1}. $$ Both terms above are devisible by $7$.

The proof is now complete if you know the induction principal.