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I was requested to show $11|3^{2n+2}+2^{6n+1}$ for $n \in \mathbb{N}$. I tried using induction. Since $n=1$ yields $81+128=209=11\cdot 19$ we can safely assume $3^{2k+2}+2^{6k+1}=11m$ for some $k \in \mathbb{N}$. Then we can consider the case for

$$3^{2(k+1)+2}+2^{6(k+1)+1}=3^{2k+4}+2^{6k+7}$$

Notice that $$ \begin{align} 3^{2k+4}+2^{6k+7} &= 3^{2k+2+2}+2^{6k+1+6} \\ &=3^{2k+2}\cdot 3^2+2^{6k+1}\cdot2^6 \end{align} $$

This seems promising, but I can't find anything to do with it. Yes, we assumed $11|3^{2k+2}+2^{6k+1}$. However, for $11$ to divide a linear combination of the two terms (as the one we find in our inductive step), we should know $11$ divides each term separately, which is not something we know.

Another despaired attempt was trying to make use of the fact that $3$, $2$ and $11$ are all primes. I know $m|n$ if and only if, for their prime factorizations $m=p_1^{e_1}p_2^{e_2}...p_k^{e_k}, n=p_1^{f_1}p_2^{f_2}...p_k^{f_k}$, we have $e_i \leq f_i$ for all $i$. However, what can we know about the prime factorization of $3^{2k+2}+2^{6k+1}$?

I'm pretty sure using induction is not the simplest way to show this (and I'm totally sure prime properties is completely desperate), and yet I have no other ideas on how to go about the problem. How can this property be demonstrated?

lafinur
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    $2^6\equiv-2(\mod11)$ and $3^2\equiv-2(\mod 11)$ – Stéphane Jaouen Nov 12 '22 at 13:32
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    It's much easier to begin with $n=0$. – José Carlos Santos Nov 12 '22 at 13:36
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    It's $,11\mid 9(9^n)+2(64^n),,$ true by $!\bmod 11!:\ 9(9^n) \equiv \color{#c00}{-2(64^n)},$ by $,\color{#c00}{-2}\equiv 9\equiv \color{#c00}{64},$ & the Congruence Product Rule. See here in the linked dupe for how to mechanically translate this modular arithmetic proof into an inductive proof, and follow the linked questions in the dupe for many more worked examples. – Bill Dubuque Nov 12 '22 at 14:58
  • You are almost done!

    For the possible final step observe that,

    $$\begin{align}3^{2k+4}+2^{6k+7}&=9\left(11m-2^{6k+1}\right)+2^{6k+7}\ &=99m-9\cdot2^{6k+1}+2^{6k+7}\ &=99m+2^{6k+1}\left(2^6-9\right)\ &=11\left(9m+5\cdot 2^{6k+1}\right).\end{align}$$

    This completes your proof.

     – lone student Nov 12 '22 at 15:03

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$3^{2n+2}+2^{6n+1}=9^{n+1}+32^n\cdot2^{n+1}=(11-2)^{n+1}+2^{n+1}(33-1)^n$ This implies $$3^{2n+2}+2^{6n+1}\equiv(-2)^{n+1}+2^{n+1}(-1)^n\pmod{11}$$ We can see that for both $n$ odd and even one has $$3^{2n+2}+2^{6n+1}\equiv0\pmod{11}$$

Piquito
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 12 '22 at 15:01
  • @Bill Dubuque: You are a respectable member of MSE and as such I respect you. My English is very poor and without the Google translator I would not be able to give answers to certain problems. In this sense I did not understand at first glance what you wanted to tell me (in particular I do not know what FAQ means). But I'll try to follow your advice after seeing all this clear. Best regards. – Piquito Nov 13 '22 at 00:40