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.
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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