How can I prove that $4^n-3n-1$ is divisible by $9$? I tried dividing the expression by $9$ and seeing if the terms cancelled in any predictable way but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it.
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2Do you know about modular arithmetic? That function is periodic mod $9$ (since both $3n$ and $4^n$ have period $3$ when taken mod $9$) so you can just check the first three values and conclude. – Milo Brandt Sep 20 '15 at 18:35
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Call $a_n=4^n-3n-1$. Then $$a_{n+1}=4^{n+1}-3n-4=4(4^n-3n-1)+9n$$ so two consecutive terms differ by a multiple of $9$. Since $a_0=0$ is divisible by $9$, all of them are.
Sonner
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again we need to prove $(4^n-3n-1)$ is divisible by 9 so that $a_{n+1}$ is also divisible by 9. Isn't that awkward ? – ViX28 Mar 29 '16 at 16:34
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Consider the equation modulo $9$. By the binomial theorem, $$4^n=(3+1)^n=\sum_{i=0}^n \binom{n}{i}3^i1^{n-i} \equiv \binom{n}{0}+3\binom{n}{1} \equiv 1+3n \pmod{9}.$$
Thus $4^n-3n-1\equiv 1+3n-3n-1=0\pmod{9}.$
jgon
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Note, $4^3 \equiv 1$ mod $9$. So you can just case on $ n\equiv 0,1,2$ mod $3$. The rest is computation.
Zach Stone
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Apply Induction on $n$
Base Case : $n=0$ . It is true.
Inductive Step: Let us assume it hold for $n = k$ i.e $$4^k -3k -1 = 9k $$ where k is an Integer. Now we need to prove
$$4^{k+1} - 3(k+1) -1 $$ is divisible by 9. $$3.4^k + 4^k -3k -1 -3$$ Now just again apply the induction two times and you are done
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Because $$4^n-3n-1=3(4^{n-1}+4^{n-2}+...+1-n)=3(4^{n-1}-1+4^{n-2}-1+...+4-1)$$ and we are done!
Michael Rozenberg
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