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

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The discussion revolves around proving that the expression 3^{2n+2} - 8n - 9 is divisible by 64 for any positive integer n. Induction is suggested as a suitable method for this proof, with a focus on the equivalence of related expressions like 9^{n+1} - 8n - 9 and 9^n - 8n - 1. The binomial expansion of 9^n is also considered, particularly in the context of simplifying the proof. A clarification is made regarding the relationship between the expressions, emphasizing that proving one implies the other is also divisible by 64. The conversation concludes with a resolution of the initial confusion about the problem.
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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 3^{2n+2} - 8n -9 is divisible by 64.
 
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I find when you're asked to prove that every member of some subset of the natural numbers has some property, induction usually works.
 
9^{n+1}-8n-9

or equivalently

9^n-8n-1

what is the binomial expansion of 9^n when considering 9=8+1?
 
Maybe I'm having a blonde moment, but doesn't a^{n + 1} - a = a(a^{n} - 1), making 9^{n + 1} - 9 - 8n = 9(9^{n} - 1) - 8n?
 
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.
 
I get it now. Thank you.
 
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