Show that $\displaystyle 3^{2n} - 8n - 1$ is divisible by 64 for integer n > 1
I suggest "proof by induction". If n= 2, we have $\displaystyle 3^4- 16- 1= 81- 17= 64$ which is obviously divisible by 64.
Now, assume that, for some k, $\displaystyle 3^{2k}- 8k- 1$ is divisible by 64. That is, that $\displaystyle 3^{2k}- 8k- 1= 64m$ for some integer m. Now look at $\displaystyle 3^{2(k+1)}- 8(k+1)-1= 9(3^{2k})- 8k -8- 1$$\displaystyle = 3^{2k}- 8k- 1- 8+ 8(3^{2k})= 64m- 8+ 8(3^{2k})= 64m- 8(3^{2k}-1)$. Now that is obviously divisible by 8 but 64? You would have to prove that $\displaystyle 3^{2k}- 1$ is divisible by 8. That is $\displaystyle 9^k- 1$. Is 1 less than a power of 9 always divisible by 8? 9- 1= 8, 81- 1= 80= 8(10), 729- 1= 728= 8(91), ...
Hmmm, looks like another proof by induction! As above it works for n= 1. Suppose that $\displaystyle 9^k- 1= 8j$ for some integers k and j. Then $\displaystyle 9^{k+1}- 1= 9(9^k)- 1= 8(9^k)+ 9^k- 1= 8(9^k)+ 8j= 8(9^k+ j)$. Yes! That completes the proof!
Hello, mark1950!
Another inductive proof . . .
Verify $\displaystyle S(1)\!:\;\;3^2 - 8(1) - 1 \:=\:0$, which is divisible by 64.Show that $\displaystyle 3^{2n} - 8n - 1$ is divisible by 64 for integer $\displaystyle n > 1$
Assume $\displaystyle S(k)\!:\;\;3^{2k} - 8k - 1 \:=\:64a\:\text{ for some integer }k.$
Add $\displaystyle 8\cdot3^{2k} - 8$ to both sides:
. . . $\displaystyle 3^{2k} {\color{blue}+ \:8\cdot 3^{2k}} - 8k\: {\color{red} -\: 8} - 1 \;=\;8a {\color{blue}\:+\: 8\cdot 3^{2k}} {\color{red}\:-\:8}$
. . $\displaystyle (8 + 1)\!\cdot\!3^{2k} - 8(k+1) - 1 \;=\;8a + 8\!\cdot\!3^{2k} - 8$
. . . . . .$\displaystyle 9\!\cdot\!3^{2k} - 8(k+1) - 1 \;=\;8a + 8\!\cdot\!3^{2k} - 8$
. . . . . $\displaystyle 3^2\!\cdot\!3^{2k} - 8(k+1) - 1 \;=\;8a + 8\!\cdot\!3^{2k} - 8 $
. . . . . .$\displaystyle 3^{2k+2} - 8(k+1) - 1 \;=\;8a + 8\!\cdot\!3^{2k} - 8 $
. . . . . $\displaystyle 3^{2(k+1)} - 8(k+1) - 1 \;=\;8\left(a + 3^{2k} - 1\right)\quad\leftarrow\:\text{ multiple of 8}$
We have proved $\displaystyle S(k+1)$.
. . The inductive proof is complete.
Very Easy.
$\displaystyle 9^n-8n-1=(1+8)^n-8n-1$
$\displaystyle
(1+8)^n=\binom{n}{0}+\binom{n}{1}8+\binom{n}{2}8^2 +\binom{n}{3}8^3+\binom{n}{4}8^4+........+\binom{n }{n}8^n
$
$\displaystyle
9^n=1+8n+8^2\left(\binom{n}{2}+\binom{n}{3}8+\bino m{n}{4}8^2+........+\binom{n}{n}8^{n-2}\right)
$
$\displaystyle
3^{2n}-8n-1=64\left(\binom{n}{2}+\binom{n}{3}8+\binom{n}{4}8 ^2+........+\binom{n}{n}8^{n-2}\right)
$
Q.E.D
P(2) is true
Assume P(k) is true or 3$\displaystyle ^{2k}-8k-1=64R$
$\displaystyle \Rightarrow 3^{2k}=64R+8K+1$-------------------(1)
$\displaystyle P(k+1)=3^{2k+2}-8k-8-1$
$\displaystyle \Rightarrow P(k+1)=3^{2k}\cdot9-8k-9$
$\displaystyle \Rightarrow P(k+1)=9(64R+8k+1)-8k-9$ [From 1]
$\displaystyle \Rightarrow P(k+1)=64R\times9+72k+9-8k-9$
$\displaystyle \Rightarrow P(k+1)=64R\times9+64k$
$\displaystyle \Rightarrow P(k+1)=64(9R+k)=64a$
So $\displaystyle 3^{2n} - 8n - 1$ is divisible by 64 for integer n > 1
Please learn something from your other thread: http://www.mathhelpforum.com/math-he...sible-7-a.html