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I cannot complete the third step of induction for this one. The assumption is $3^{2n+3}+40n-27=64k$, and when substituting for $n+1$ I obtain $3^{2n+5}+40n+13=64k$. I've tried factoring the expression, dividing, etc. but I cannot advance from here.

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First, show that this is true for $n=1$:

$3^{2+3}+40-27=64\cdot4$

Second, assume that this is true for $n$:

$3^{2n+3}+40n-27=64k$

Third, prove that this is true for $n+1$:

$3^{2(n+1)+3}+40(n+1)-27=$

$9\cdot(\color\red{3^{2n+3}+40n-27})-320n+256=$

$9\cdot\color\red{64k}-320n+256=$

$576k-320n+256=$

$64\cdot(9k-5n+4)$


Please note that the assumption is used only in the part marked red.

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    $\begingroup$ This is a special case of a two-term truncation of the Binomial Theorem - see my answer. $\endgroup$ – Gone Jul 6 '16 at 15:41
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Many inductive proofs of this and similar divisibilities boil down to computing the first two terms of the Binomial Theorem $\,\color{#0a0}{\rm BT}.\,$ However, this innate structure is often (greatly) obfuscated by details of the special case. Let's bring this structure to the fore and exploit it to the hilt. Doing so we will see how it greatly simplifies the inductive proof - so much so that the proof becomes obvious, and the arithmetic becomes so easy that it can all be done mentally.

$ {\rm mod}\,\ \color{#c00}{8^2}\!:\,\ \ \overbrace{27 (1\!+\!8)^n}^{\large 3^{\LARGE 3\,+\,2n}} \,\overset{\color{#0a0}{\rm\large BT_{\phantom |}\!}}\equiv {27(1\!+\!8n)} \equiv \!\!\!\!\!\!\!\!\overbrace{27\!+\!24n}^{\large -(-27\,+\,40n)\ \ \quad }\!\!\!\!\! $ by $\,\ 27(8)\equiv \overbrace{{(3\!+\!3\!\cdot\! \color{#c00}8)}\color{#c00}8 \equiv 3\cdot8}^{\large \color{#c00}{8^{\Large 2}}\,\equiv\, 0}\,$

The inductive step of $\,\color{#0a0}{\rm BT}\,$ is much clearer without obfuscation by special-case cruft, viz.

$\!\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\ \ (1+ a)^{\large n}\ \ \ \ \equiv&\,\ \ 1 + na\qquad\qquad\ \ \ \ \,{\rm i.e.}\ \ P(n)\\[1pt] \Rightarrow\ \ \ (1+a)^{\color{}{\large n+1}}\! \equiv &\ (1+na)(1 + a)\ \ \ \ \ \ \ {\rm by}\ \ 1+a \ \ \rm times\ prior\\[2pt] \equiv &\,\ \ 1+ na+a+n\,\color{#c00}{a^2}\\[2pt] \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\quad\ \, {\rm i.e.}\ \ P(\color{}{n\!+\!1})\\ \end{align}$

More generally the same idea yields a proof of the following

Theorem $\ \ \forall n\in\Bbb N\!:\ d\mid f(n) = a^n\! + bn + c \iff d\mid \color{}{(a\!-\!1)^2},\, \color{}{a\!+\!b\!-\!1},\, \color{}{1\!+\!c}$

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  • $\begingroup$ Nice answer, but what does cruft mean? $\endgroup$ – J. W. Tanner Sep 12 '19 at 12:25
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    $\begingroup$ @J.W.T See here for "cruft", which above refers to special-case details that obscure the general essence of the matter (not much above but it can be worse in more complicated instances). I added a link to a more general result which clarifies the role of the Binomial Theorem. $\endgroup$ – Gone Sep 12 '19 at 14:03
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HINT: $$\begin{align} & 3^{2n+5}+40n+13 \\ &=9\cdot 3^{2n+3}+9\cdot 40n -9\cdot 27 - 8\cdot 40n+256 \\ & =9(3^{2n+3}+40n -27) - 64\cdot 5n+ 64\cdot 4\end{align}$$

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Notice that: $$3^{2n+5}+40n+13=9(3^{2n+3}+40n−27)−320n+256.$$

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Let $A_n = 3^{2n+3}+40n-27$, then $A_n = 11A_{n-1} - 19A_{n-2} + 9A_{n-3}$.

From this it's clear that if 64 divides $A_n$ three consecutive values of $n$ then it holds for the next. So by induction it's enough to check it for $n = -1,0,1$, which is easy enough to do by hand.

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  • $\begingroup$ You probably mean $n=0, 1, 2$. $\endgroup$ – Catalin Zara Jul 6 '16 at 14:54
  • $\begingroup$ @CatalinZara Any three consecutive values work, and it's easier to check for $-1,0,1$ $\endgroup$ – πr8 Jul 6 '16 at 15:33
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    $\begingroup$ In case anyone wonders where the recurrence comes from it is $\,(S-9)(S-1)^2 A_n = 0,\,$ where $\,S\,$ is the shift operator $\,S A_n = A_{n+1}.\ $ Here $\,S-9\,$ kills $\,3^{2n+2} = 27\cdot 9^n,\,$ and $\,S-1\,$ kills $\,40n-27.\,$ since $\,S-1\,$ decreases degree by $1$. Further, the factors $\,S-c_i\,$ commute since they have coef's $\,c\,$ that are constant, i.e. $\,S c = c.\ $ $\endgroup$ – Gone Jul 6 '16 at 16:02

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