Prove that $3^{4n-1}-2$ is always a multiple of 5.

Question

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?

Answer 1

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 $81$.

Answer 2

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 $$a-b|a^n-b^n$$

Answer 3

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 $$3^{4n-1}-2=3^{4n}3^{-1}-2\equiv 1\cdot 2-2=0\bmod 5.$$

Answer 4

$\displaystyle 3^{4n-1}=\frac{(3^4)^n}{3}=\frac{81^n}{3}=27\cdot 81\cdot 81\cdot ...$

The last digit is always $7$, so $3^{4n-1}-2$ ends with $5$

Answer 5

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.

Answer 6

$3^{4n-1}-5+3= 3(3^{4n-2}+1)-5=$

$3((3^2)^{2n-1}+1)-5=$

$3((10-1)^{2n-1}+1)-5=$

$3(\sum_{k=0}^{2n-2}\binom{2n-1}{k}10^{(2n-1)-k}(-1)^k +(-1)^{2n+2}-1) -5.$

All terms in the binomial expansion, except the last (+1), have the factor 10.