Prove that $5^{2n-1} - 3^{2n-1} - 2^{2n-1}$ is divisible by 15 for n $\in$ $\mathbb{N}$

Question

The book I am using for my Combinatorics course is Combinatorics:Topics, Techniques, and Algorithms.

Prove that $5^{2n-1} - 3^{2n-1} - 2^{2n-1}$ is divisible by 15 for n $\in$ $\mathbb{N}$

This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin:

Proof:

Answer 1

Without induction: we want to prove $A=5^{2n-1}-3^{2n-1}-2^{2n-1}$ is divisible by $3$ and $5$.

$A=5^{2n-1}-\left(3^{2n-1}+2^{2n-1}\right)=5\left(5^{2n-2}-\left(3^{2n-2}-3^{2n-3}2\pm\cdots + 2^{2n-2}\right)\right)$.

$A=\left(5^{2n-1}-2^{2n-1}\right)-3^{2n-1}=3\left(\left(5^{2n-2}+5^{2n-3}2+\cdots + 2^{2n-2}\right)-3^{2n-2}\right)$.

Answer 2

You can also do it like this.

For divisibility by $5$ it is clear that $5^{2n-1}$ is divisible by $5$ and $$3^{2n-1}+2^{2n-1}=(3+2)(3^{2n-2}-2\cdot 3^{2n-3}+2^2\cdot 3^{2n-4}-\dots +2^{2n-2})$$ is clearly an integer multiple of $5$

Also $5^r-2^r=(5-2)(5^{r-1}+2\cdot 5^{r-2}+ \dots +2^{r-1})$ is divisible by $3$ for any $r$ and not just odd values. So you can easily show divisibility by $3$.

Answer 3

Base case: $5-3-2$ is divisible by $15$.

Assume that $5^{2k-1}-3^{2k-1}-2^{2k-1}=15m$ for some integer $m$.

Then $5^{2k+1}-3^{2k+1}-2^{2k+1}=25\cdot5^{2k-1}-9\cdot3^{2k-1}-4\cdot 2^{2k-1}$

$=21\cdot 5^{2k-1}-5\cdot 3^{2k-1}+4(15m)$

Now each term has a factor of a $5$ and a $3$, so the number is divisible by $15$.