Using Binomial Theorem, prove that $9^{n+1} - 8n - 8$ always leaves remainder 1 when divided by 64, whenever $n$ is a positive integer.

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> [!TLDR]
> Expanding $9^{n+1}=(1+8)^{n+1}$ by the Binomial Theorem, every term from the third one onward is divisible by $64$; collecting the first two terms separately shows $9^{n+1}-8n-9$ is exactly divisible by $64$, so $9^{n+1}-8n-8$ leaves remainder $1$.
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> Write $9 = 1+8$, so $9^{n+1} = (1+8)^{n+1}$. By the Binomial Theorem:
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> $(1+8)^{n+1} = \displaystyle\sum_{r=0}^{n+1}\binom{n+1}{r}8^r = \binom{n+1}{0} + \binom{n+1}{1}8 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3+\cdots+8^{n+1}$
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> $= 1 + 8(n+1) + 8^2\left[\binom{n+1}{2}+\binom{n+1}{3}8+\cdots+8^{n-1}\right]$
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> Every term from the third onward carries a factor of $8^2=64$, so group them as $64k$ for some non-negative integer $k$ (the bracketed sum is always a non-negative integer):
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> $9^{n+1} = 1+8(n+1)+64k = 1+8n+8+64k = 8n+9+64k$
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> Rearranging:
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> $9^{n+1}-8n-9 = 64k$
> …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.Using Binomial Theorem, prove that $9^{n+1} - 8n - 8$ always leaves remainder 1 when divided by 64, whenever $n$ is a positive integer.

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

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Q.Using Binomial Theorem, prove that 9n+1−8n−89^{n+1} - 8n - 89n+1−8n−8 always leaves remainder 1 when divided by 64, whenever nnn is a positive integer.

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Q.Using Binomial Theorem, prove that $9^{n+1} - 8n - 8$ always leaves remainder 1 when divided by 64, whenever $n$ is a positive integer.

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