Divisibility result by induction: $64 \mid 3^{2n + 2} - 8n - 9$

Question

Prove by induction that:

$64 \mid 3^{2n + 2} - 8n - 9$, $n > 0$

I've tried to manipulate $P(k + 1)$ and haven't come up with anything like $P(k)$.

$P(k): 64 \mid 3^{2k + 2} - 8k - 9, k > 0 $
$P(k+1): 64 \mid 3^{2(k+1) + 2} - 8(k+1) - 9$

Answer 1

Note that\begin{align}3^{2(k+1)+2}-8(k+1)-9&=9\times3^{2k+2}-8k-9-8\\&=8(3^{2k+2}-1)+3^{2k+2}-8k-9.\end{align}Now, you are assuming that $64\mid3^{2k+2}-8k-9$. So, all that remains to be proved is that $64\mid8(3^{2k+2}-1)$, which is equivalent to the assertion that $8\mid3^{2k+2}-1$. But that is easy, since $3^{2k+2}=9^{k+1}=(8+1)^{k+1}$.

Answer 2

If $f(m)=3^{2n+2}-8n-9,$

Eliminate $3^{2n}$ to establish $$f(m+1)-3^2f(m)$$ is actually divisible by $64$

$$\implies64\mid f(m+1)\iff64\mid9f(m)$$

Alternatively

If induction is not mandatory, we can use a more straight forward way using binomial expansion

$$3^{2n+2}=(1+8)^{n+1}\equiv1+\binom{n+1}18\pmod{8^2}$$

Answer 3

You know $64\mid3^{2n+2} -8n -9$.

You want to show $64\mid3^{2(n+1) + 2} - 8(n+1) - 9$

So it suffices to show that $3^{2n + 4} - 8(n+1) - 9 = (3^{n+2} -8n -9) + 64M$ for some integer $M$.

So what is $(3^{2n+4} - 8(n+1) - 9) - (3^{2n+2} -8n -9)$?

$$(3^{2n+4} - 8(n+1) - 9) - (3^{2n+2} -8n -9) = (3^{2n+4}-3^{2n+2})-(8(n+1)-8n) =$$

$$3^{2n+2}(3^2-1) -8=8*3^{2n+2}-8=8(3^{2n+2} - 1)=$$

$$8(3^2 -1)(3^{2n} + 3^{2n-2}+ 3^{2n-4} + ...... + 3^2 + 1)=$$

$$8*8(3^{2n} + 3^{2n-2}+ 3^{2n-4} + ...... + 3^2 + 1)=$$

$$64(3^{2n} + 3^{2n-2}+ 3^{2n-4} + ...... + 3^2 + 1)$$

Answer 4

$f(k) =3^{2k + 2} - 8k - 9 $

$f(0) =3^2-9 =0 $

$\begin{array}\\ f(k+1)-f(k) &=3^{2(k+1) + 2} - 8(k+1) - 9-(3^{2k + 2} - 8k - 9)\\ &=3^{2k+4}-3^{2k+2}-8\\ &=3^{2k+2}(3^2-1)-8\\ &=8(3^{2k+2}-1)\\ &=8(9^{k+1}-1)\\ &=8(9-1)\sum_{j=0}^k 9^j\\ &=64\sum_{j=0}^k 9^j\\ \end{array} $

so $64 \mid f(k)$ for all $k$.