0

Prove that $9^{n+1} - 8n- 9$ divisible by $64$.

I was wondering if the same can be proved by congruency modulus. It can be done using induction or binomial theorem.

Can anyone prove it using congruences if possible?

Vidyanshu Mishra
  • 10,391
Siva Sankar Sajeev
  • 69
  • Always remember, these kind of problems can be solved in three elementary ways. First induction second binomial theorem and third congruences but the core idea lies in congruences always. – Vidyanshu Mishra Nov 01 '17 at 22:40

3 Answers3

3

\begin{align}9^{n+1}-8n-9&=(8+1)^{n+1}-8n-9\\&=8^{n+1}-\binom{n+1}18^n+\cdots+\binom{n+1}{n-1}8^2+\overbrace{\binom{n+1}n}^{=n+1}8+1-8n-9\\&=8^{n+1}-\binom{n+1}18^n+\cdots+\binom{n+1}{n-1}8^2\\&\equiv0\pmod{64}.\end{align}

José Carlos Santos
  • 440,053
2

$9^{n+1}-8n-9$ is divided by $8$ since $9=1$ mod $8$, we have $9^{n+1}-8n-9=9(9^n-1)-8n=9(9-1)(1+...+9^{n-1})-8n=9(8.(1+...+9^{n-1}))-8n$. We have ${{9^{n+1}-8n-9}\over 8}=9(1+9+...+9^{n-1})-n$ is divided by $8$. This implies the result since it shows that $9^{n+1}-8n-9$ is divided by $8^2=64$.

Tsemo Aristide
  • 89,587
1

proof that $9^{n+1}-8n-9$ is always divisible by $64$ for all real positive values of $n$.
say $9^{n+1}-8n-9$ is a multiple of $x$, where $x$ is also a real positive integer.
$$ f_{n} = 9^{n+1}-8n-9 $$.
therefore.
$$ \begin{align} f_{n+1} = 9^{n+2}-8(n+1)-9\\ f_{n+1} = 9^{n+1}×9-8n-8-9\\ \frac{f_{n+1}}{9} = 9^{n+1}-\frac{8n}{9}-\frac{17}{9}\\ \frac{f_{n+1}}{9}-\frac{64n}{9}-\frac{64}{9} = 9^{n+1}-8n-9\\ \frac{f_{n+1}}{9}-\frac{64n}{9}-\frac{64}{9} = f_{n}\\ f_{n+1}-64n-64 = 9*f_{n}\\ f_{n+1}-9*f_{n} = 64n+64\\ \text{divide by 64}\\ \frac{f_{n+1}-9f_{n}}{64} = n+1 \end{align} $$.
since $n$ is a real positive integer, $f_{n}$ and $f={n+1}$ must also be a real positive integer.
therefore both $f_{n}$ and $f_{n+1}$ must be divisible by $64$ for $n+1$ to be a real positive integer.
$\frac{f_{n+1}-9*f_{n}}{64} = n+1$

Aderinsola Joshua
  • 1,070