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I have been trying to complete this proof by induction and am stuck. Could someone please help me? Here is my working out.

Prove that $p(n): 9^{n+1}-8n-9$ is divisible by 64 where n is a natural number using proof by induction.

Step 1: Show that $p(1)$ is true \begin{align*} p(1) & = 9^2-8-9 \\ & = 81-17 \\ & = 64 \end{align*} $64$ is divisible by $64$ so $p(1)$ is true.

Step 2: Show that $p(k)=>p(k+1)$ is true.

Assume $9^{k+1}-8k-9$ is divisible by $64$. So, $$9^{k+1}-8k-9=64M$$ where M is a natural number.

Show $p(k+1)$ is divisible by $64$. \begin{align*} 9^{k+2}-8(k+1)-9 & = 9\times9^{k+1}-8k-8-9 \\ & = (8+1)9^{k+1}-8k-9-8 \\ & = 8\times9^{k+1}+(9^{k+1}-8k-8)-9 \\ & = 64M + 8\times9^{k+1} - 8 \end{align*}

This is clearly not divisible by $64$, what have I done wrong or what am I missing?

Thank you.

N. F. Taussig
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gazelle51
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2 Answers2

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Clearly, the assertion is true for $n = 1$.

$p(1)$ is true because $$ 9^{n + 1} - 8 n - 9 = 9^2 - 8 - 9 = 64 $$ which is divisible by 64.

Induction step: Suppose that $p(n)$ is true for some positive integer $n$.

Then $$ 9^{n + 1} - 8 n - 9 = 64 M, $$ where $M$ is a positive integer.

To show that $p(n + 1)$ is true, we proceed as follows.

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

We wish to prove that $T$ is divisible by 64.

Now, we rewrite $T$ as $$ T = 9^{n + 1} \times 9 - 8 n - 8 - 9 = (8 + 1) 9^{n + 1} - 8 n - 8 - 9 $$

Thus, $$ T = 8 \left( 9^{n + 1} \right) + 9^{n + 1} - 8 n - 8 - 9 $$

Hence, we find that $$ T = 8 \left( 9^{n + 1} - 1 \right) + [9^{n + 1} - 8 n - 9] = 8 \left( 9^{n + 1} - 1 \right) + 64 M $$

By binomial expansion, $$ 9^{n + 1} - 1 = (8 + 1)^{n + 1} - 1 = \left[ 8^{n + 1} + \left( \begin{array}{c} n + 1 \\ 1 \end{array} \right) 8^n + \cdots + \left( \begin{array}{c} n + 1 \\ n \end{array} \right) 8 + 1 \right] - 1 $$

Thus, $$ 9^{n + 1} - 1 = 8 L,$$ where $L$ is some positive integer.

Hence, $$ T = 64 L + 64 M = 64 (L + M) $$

Thus, we conclude that $p(n + 1)$ is true.

Using the principle of mathematical induction, we conclude that $p(n)$ is true for all positive integers $n$. $\ \ \ \ \ \blacksquare$

Dr. Sundar
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After you simplify to $64M + 8 \cdot 9^{k+1} - 8$, we can simplify further by noticing $$64M + 8 \cdot 9^{k+1} - 8 = 64M + 8(9^{k+1} - 1)$$ so it suffices to show that $9^{k+1} - 1$ is divisible by $8$. However, $$9^{k+1} - 1 = (8+1)^{k+1} - 1 = -1 + \sum_{i=0}^{k+1}\binom{k+1}{i}8^i = \sum_{i=1}^{k+1}\binom{k+1}{i}8^i = 8\sum_{i=1}^{k+1}\binom{k+1}{i}8^{i-1}$$ which is divisible by $8$.

PTrivedi
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