Prove that $3^{4n-1}-2$ is always a multiple of 5. I am trying to prove the following statement: $f(n) =3^{4n-1}-2 $ is always a multiple of 5, for $n\in \mathbb Z^+$. Using proof by induction:
Base case: $f(1)=25$, which is a multiple of 5 and hence holds for $n=1$.
Assumption step: Assume $f(k) =3^{4k-1}-2 $ is a multiple of of 5.
Considering $n= k+1$: $f(k+1)=3^{4k+3}-2 $ 
$$=27(3^{4k})-2$$
$$f(k+1)-6f(k)^{**}=27(3^{4k})-2-6(3^{4k-1}-2) $$
$$=27(3^{4k})-2-2(3^{4k}-6)$$
$$=27(3^{4k})-2-2(3^{4k})+12$$
$$=25(3^{4k})+10$$
$$=5(5(3^{4k})+2)$$
**when performing this step, is any multiple of the $n = k$ case allowed to be added or subtracted, or is the arguement not valid for certain values?
 A: $\displaystyle 3^{4n-1}=\frac{(3^4)^n}{3}=\frac{81^n}{3}=27\cdot 81\cdot 81\cdot ...$
The last digit is always $7$, so $3^{4n-1}-2$ ends with $5$
A: It is obvious using lil' Fermat:
As $3$ is coprime to $5$, we have $3^4\equiv  1\bmod 5$. Further, its inverse $\bmod 5$ is $2$, so
$$3^{4n-1}-2=3^{4n}3^{-1}-2\equiv 1\cdot 2-2=0\bmod 5.$$
A: Yes, you can prove a quantity's divisible by $5$ by showing it to be any linear combination, with integer coefficients, of known multiples of $5$, in this case $6f(k)+5(5\cdot 3^{4k}+2)$. An easier way to do it is to compute $f(k+1)-3^4f(k)=2(3^4-1)=160$, as this isn't $k$-dependent. In other words, your coefficient of $6$ needs more work than using $81$.
A: So we need to prove that \begin{equation}
5 | 3^{4 n-1}-2 \quad \forall n \in \mathbb Z^{+}
\end{equation}
For n=1,  we have \begin{equation}
5 | 25
\end{equation}. We assume it is true for some $n\in\mathbb Z^+$
Now we prove \begin{equation}
\begin{array}{l}
{3^{4(n+1)-1}-2} \\
{3^{4 n+4-1}-2} \\
\end{array}
\end{equation}
$3^{4} \cdot\left(3^{4 n-1}\right)-2$
\begin{equation}
\begin{aligned}
&3^{4}\left(3^{4 n-1}-2\right)+2 \cdot 3^{4}-2\\
&3^{4}\left(3^{4 n-1}-2\right)+160
\end{aligned}
\end{equation}
\begin{equation}
3^{4 n-1}-2
\end{equation}  can be divided by 5 and 160 divided by 5 is 32 so whole is expression can be divided by 5.
A: An easier way
$$5|3^{4n-1}-2\overset{\gcd(5,3)=1}{\iff} 5|3^{4n}-6\iff 5|3^{4n}-1\iff 5|81^{n}-1$$and note that $$a-b|a^n-b^n$$
A: $3^{4n-1}-5+3= 3(3^{4n-2}+1)-5=$
$3((3^2)^{2n-1}+1)-5=$
$3((10-1)^{2n-1}+1)-5=$
$3(\sum_{k=0}^{2n-2}\binom{2n-1}{k}10^{(2n-1)-k}(-1)^k +(-1)^{2n+2}-1) -5.$
All terms in the binomial expansion, except the last (+1), have  the factor 10.
