# Prove 5|(3^(3n+1)+2^(n+1)) for every positive integer n.

• Tsunoyukami
In summary, the expression ##3^{3(k+1)+1}+2^{(k+1)+1}## can be written as ##25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##, and since ##5|(3^{3k+1} + 2^{k+1})##, we can write it as ##5(5\cdot3^{3k+1}+2x)##, which is divisible by 5. Therefore, ##5|(3^{3(k+1)+1}+2^{(k+1)+1})## as well.
Tsunoyukami
"Prove ##5|(3^{3n+1}+2^{n+1})## for every positive integer ##n##." (Exercise 11.8 from Mathematical Proofs: A Transition to Advanced Mathematics 3rd edition by Chartrand, Polimeni & Zhang; pg. 282)

I'm having difficulty solving this exercise. My first thought was to use induction but I get stuck with an expression that I can't seem to manipulate into the required form. The next thing I tried was considering cases (when ##n## is even and when ##n## is odd) but once again I get stuck.

Can anyone suggest another method of proof?

I feel that induction should work out for me despite my difficulties. As such, I will explain what I've done so far. If you would like to see work I describe but do not include below let me know and I'll include it.

First I showed that this expression holds for the case ##n=1##. I assumed this holds for some arbitrary positive integer ##k##. Now I wish to show that ##5|(3^{3(k+1)+1}+2^{(k+1)+1})##.

From here I tried working with the expression ##3^{3(k+1)+1}+2^{(k+1)+1}##:

##3^{3(k+1)+1}+2^{(k+1)+1}=3^{3k+3+1}+2^{k+1+1}=3^{3k+4}+2^{k+2}=3^{3k} \cdot 3^4 + 2^{k} \cdot 2^{2}=81\cdot3^{3k}+4\cdot2^{k}##

This is where I get stuck - I don't know how to massage this equation to show that 5 divides it.

Tsunoyukami said:
"Prove ##5|(3^{3n+1}+2^{n+1})## for every positive integer ##n##." (Exercise 11.8 from Mathematical Proofs: A Transition to Advanced Mathematics 3rd edition by Chartrand, Polimeni & Zhang; pg. 282)

I'm having difficulty solving this exercise. My first thought was to use induction but I get stuck with an expression that I can't seem to manipulate into the required form. The next thing I tried was considering cases (when ##n## is even and when ##n## is odd) but once again I get stuck.

Can anyone suggest another method of proof?

I feel that induction should work out for me despite my difficulties. As such, I will explain what I've done so far. If you would like to see work I describe but do not include below let me know and I'll include it.

First I showed that this expression holds for the case ##n=1##. I assumed this holds for some arbitrary positive integer ##k##. Now I wish to show that ##5|(3^{3(k+1)+1}+2^{(k+1)+1})##.

From here I tried working with the expression ##3^{3(k+1)+1}+2^{(k+1)+1}##:

##3^{3(k+1)+1}+2^{(k+1)+1}=3^{3k+3+1}+2^{k+1+1}=3^{3k+4}+2^{k+2}=3^{3k} \cdot 3^4 + 2^{k} \cdot 2^{2}=81\cdot3^{3k}+4\cdot2^{k}##

This is where I get stuck - I don't know how to massage this equation to show that 5 divides it.

Write that as ##3^{3(k+1)+1}+2^{(k+1)+1}=27*3^{3k+1}+2^{k+1}##. Note 27=25+2.

Thanks so much for the hint!

So we can write:

##3^{3(k+1)+1}+2^{(k+1)+1}##
##= 27 \cdot 3^{3k+1}+2\cdot 2^{k+1}##
##=(25+2)\cdot 3^{3k+1}+2 \cdot 2^{k+1}##
##=25 \cdot 3^{3k+1} + 2 \cdot 3^{3k+1} + 2\cdot 2^{k+1}##
##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##

By assumption we know that ##5|(3^{3k+1} + 2^{k+1})## so we can write ##(3^{3k+1} + 2^{k+1})=5x## for some integer x.

Then we have
##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##
##=5(5\cdot3^{3k+1})+5\cdot2x##
##=5(5\cdot3^{3k+1}+2x)##

Since ##5\cdot3^{3k+1}+2x## is an integer, we conclude that ##5|3^{3(k+1)+1}+2^{(k+1)+1}##.

Last edited:
Tsunoyukami said:
Thanks so much for the hint!

So we can write:

##3^{3(k+1)+1}+2^{(k+1)+1}##
##= 27 \cdot 3^{3k+1}+2\cdot 2^{k+1}##
##=(25+2)\cdot 3^{3k+1}+2 \cdot 2^{k+1}##
##=25 \cdot 3^{3k+1} + 2 \cdot 3^{3k+1} + 2\cdot 2^{k+1}##
##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##

By assumption we know that ##5|(3^{3k+1} + 2^{k+1})## so we can write ##(3^{3k+1} + 2^{k+1})=5x## for some integer x.

Then we have
##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##
##=5(5\cdot3^{3k+1})+5\cdot2x##
##=5(5\cdot3^{3k+1}+2x)##

Since ##5\cdot3^{3k+1}+2x## is an integer, we conclude that ##5|3^{3(k+1)+1}+2^{(k+1)+1}##.

Right. You definitely want to keep ##3^{3k+1} + 2^{k+1}## in the answer somehow because you know it's divisible by 5. And splitting the factor up becomes pretty obvious once you've seen the trick.

## 1. What does the expression "5|(3^(3n+1)+2^(n+1))" mean?

The expression means that 5 is a factor of the quantity (3 to the power of 3n+1) plus (2 to the power of n+1).

## 2. How do you prove that 5 is a factor of the given expression?

This can be proven using mathematical induction. We can show that the expression is divisible by 5 for the base case (n=1) and then use the inductive hypothesis to prove that it is also divisible for n+1.

## 3. Why is it important to prove this statement for every positive integer n?

Proving it for every positive integer n ensures that the statement is true for all possible values of n, rather than just a few specific values. This makes the proof more rigorous and reliable.

## 4. Can you provide an example to illustrate this statement?

Sure, for n=2, the expression becomes 5|(3^(3*2+1)+2^(2+1)) or 5|(3^7+2^3). We can divide both terms by 5 and get 5|(2187+8), which is clearly true.

## 5. What is the significance of this statement in mathematics or real-world applications?

This statement is significant because it helps us understand the divisibility rules and properties of numbers. It also has applications in number theory and cryptography. Additionally, proving statements like this can help us develop problem-solving skills and logical reasoning.

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