Solving Problem: Prove 3^{2n+2} - 8n -9 Divisible by 64

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In summary, the conversation discusses how to prove that the expression 3^{2n+2} - 8n -9 is divisible by 64 for any positive integer n. The use of induction is suggested, and the binomial expansion of 9^n is considered. It is noted that proving one expression is divisible by 64 will also mean the other expression is divisible by 64. Finally, it is clarified that the expressions are not equal, but proving one is divisible by 64 will also prove the other to be divisible by 64.
  • #1
recon
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Can someone point me in the right direction of solving the following problem:

Prove that for any postive integer n, the value of the expression [tex]3^{2n+2} - 8n -9[/tex] is divisible by 64.
 
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  • #2
I find when you're asked to prove that every member of some subset of the natural numbers has some property, induction usually works.
 
  • #3
9^{n+1}-8n-9

or equivalently

9^n-8n-1

what is the binomial expansion of 9^n when considering 9=8+1?
 
  • #4
Maybe I'm having a blonde moment, but doesn't [itex]a^{n + 1} - a = a(a^{n} - 1)[/itex], making [itex]9^{n + 1} - 9 - 8n = 9(9^{n} - 1) - 8n[/itex]?
 
  • #5
ah, perhaps i ought to have been clearer: i wasn't say the epxressions are equal, but that if you prove one is divisible by 64 for all n, the other will be divislbe by 64 for all n (give or take a case when n=0). I let m=n+1 in the first, then relabelled n=m.
 
  • #6
I get it now. Thank you.
 

Related to Solving Problem: Prove 3^{2n+2} - 8n -9 Divisible by 64

1. How do I approach solving this problem?

The best approach to solving this problem is by using mathematical induction. This involves proving the statement for a base case, typically n=0, and then using the assumption that it holds for n=k to prove it holds for n=k+1.

2. What is the base case for this problem?

The base case for this problem is when n=0. When n=0, the statement becomes 3^{2(0)+2} - 8(0) - 9 = 3^2 - 9 = 0 which is divisible by 64.

3. How do I prove that the statement holds for n=k+1?

To prove that the statement holds for n=k+1, you need to show that if the statement holds for n=k, then it also holds for n=k+1. This can be done by substituting n=k+1 into the original statement and using the assumption that it holds for n=k to simplify the expression.

4. How do I show that the statement is true for all positive integers?

To show that the statement is true for all positive integers, you need to prove it for the base case (n=0) and then show that if it holds for n=k, it also holds for n=k+1. This will prove that the statement holds for all positive integers through mathematical induction.

5. Can this problem be solved using other methods?

Yes, there are other methods that can be used to solve this problem such as using the rules of divisibility or using modular arithmetic. However, using mathematical induction is the most efficient and reliable method for solving this type of problem.

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