Proofing divisibility by $7$

Question

Assignment: Proof by induction that $7|(2^{n+1}+3^{2k-1})$ for all $n\geq 1$

I skip the test with $n=1$

let $n=k$ assume it's true for some $k$, then

$a\in N$

$$2^{k+1}+3^{2k-1}=7a$$

$$2^{k+1}=7a-3^{2k-1}$$

Let $n=k+1$

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

Insert assumption:

$$=2(7a-3^{2k-1})+3^{2k+1}$$

$$=14a-2(3^{2k-1})+3^{2k+1}$$

Since $14$ is divisible by $7$ it can be dropped of the equasion. (but how to do it rigorous?) \begin{align} 2(-3^{2k-1})+3^{2k+1}& =(-3^{2k})\frac{2}{3}+3(3^{2k})\\ &=3^{2k}(3-\frac{2}{3})\\ &=3^{2k}(\frac{7}{3})\\ &=7(\frac{3^{2k}}{3}) \end{align} (what to say here?)

Answer 1

Your last expression is $7(3^{2k-1})$. Since $k\geq 1$, $3^{2k-1}$ is an integer.

To make things more rigourous, you can keep the term or use the notation

$$ 14a-2(3^{2k-1})+3^{2k+1} \equiv -2(3^{2k-1})+3^{2k+1} \mod 7$$

or you can just keep the terms there in every equation.

Answer 2

When you are at $14a-3^{2k-1}\cdot2+3^{2k+1}$ go on as follows: $$ 14a-3^{2k-1}\cdot2+3^{2k+1}= 14a-3^{2k-1}\cdot2+3^{2k-1}\cdot3^2= 14a-3^{2k-1}(2-9)=\dots $$