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Question as in the title : does anyone know how to prove that $3^n$ does not divide $8^n+1$ for $n\geq 4$ or find a counterexample ?

My thoughts : I have checked that this is true for $n\leq 1000$. One can easily show that certain congruence classes are excluded : for example if $n$ is even, then $8^n+1$ is congruent to $2$ modulo $3$ and so it is not divisible by $3$, if $n$ is congruent to $5$ modulo $6$ then $8^n+1$ is congruent to $18$ modulo $27$ and so it is not divisible by $27$, etc.

On the other hand, it is equally easy to show that $8^n+1$ can be made divisible by arbitrarily large powers of $3$, so I'm not sure that the congruence method helps.

Ewan Delanoy
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2 Answers2

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Define

$\mathbb v_p(n)$ : The $p$-adic order (valuation) of number $n$ is the number of times $p$ divides $n$.

Then,

You want to prove that $\mathbb v_3(8^n+1)\lt n$ for all $n\ge 4$.

Notice that:

$$ \mathbb v_3(8^n+1)=\begin{cases} 0, & n\text{ even}\\ 2, & n\equiv1,5\pmod{6}\\ 3, & n\equiv3,15\pmod{18}\\ 4, & n\equiv9,45\pmod{54}\\ \dots\\ k, & n\equiv3^{k-2},5\cdot 3^{k-2}\pmod{2\cdot 3^{k-1}}\\ \dots \end{cases} $$

That is, $\mathbb v_3(8^n+1) = k$ for the first time when $n=3^{k-2}$. Hence for $k=n$,

$$ n\lt 3^{n-2} \implies \mathbb v_3(8^n+1) \lt n $$

It is easy to see that LHS holds if and only if $n\ge 4$.

Q.E.D.

Vepir
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  • Thanks for the feedback. Can you prove that $\mathbb v_3(8^n+1)=k$ for $n=3^{k-2}$ or is it just a guess ? – Ewan Delanoy Apr 15 '20 at 14:42
  • Ah I'm starting to see the pattern : $8^n+1=\frac{8^{2n}-1}{8-1}$, so we just need to compute $\mathbb v_3(8^{2n}-1)$. Also by Fermat's small theorem we have $8^{6n} \equiv 8 (mod\ 6n)$. – Ewan Delanoy Apr 15 '20 at 14:49
  • @EwanDelanoy The formula for $\mathbf v_3(8^n+1)$ is based on the generalizing the pattern of what you already observed in your post under "My thoughts". I hope this helps. – Vepir Apr 15 '20 at 14:52
  • It sure helps. Time will tell if it easily can be turned into a full proof – Ewan Delanoy Apr 15 '20 at 14:53
  • @EwanDelanoy Alternatively, you are looking for a rigorous proof that shows: $$ \mathbf v_3(2^n+1) =\begin{cases} \mathbf v_3(n) + 1, & n\text{ odd}\ 0& n\text{ even} \end{cases}$$

    Because then the proof is easy: (last "$\lt$" is for $n\ge 4$) $$\mathbf v_3(8^n+1) = \mathbf v_3(2^{3n}+1) \le \mathbf v_3(3n) + 1 = \mathbf v_3(n) + 2\le\log_3n+2\lt n. \square$$

     – Vepir Apr 16 '20 at 12:29
  • ${\mathbb v}_3(2^{2n+1}+1)={\mathbb v}_3(n)+1$ is indeed true for $1\leq n \leq 1000$. This looks very promising also ! – Ewan Delanoy Apr 16 '20 at 17:37
  • @EwanDelanoy I was unable to turn this observation into a rigorous proof, hence I've presented it in a separate question combining the details that we've discussed so far. – Vepir Apr 16 '20 at 21:08
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From the start we can reason that $n$ must be odd since when it's even it's never divisible. $$8^n+1 \equiv (9-1)^n +1 \equiv (-1)^n+1 \equiv 2 \mod 3$$

Now that we know $n$ is odd, we can use the lifting the exponent lemma (LTE), because,

$$v_3(8^n+1) = v_3(8^n-(-1)^n)$$

So we check the criteria for the LTE $$v_3(8) = v_3(-1) = 0$$ $$v_3(8-(-1))\ge 1$$

So we have,

$$v_3(8^n-(-1)^n) = v_3(n) + v_3(8-(-1))$$

$$v_3(8^n+1) = v_3(n)+2$$

Because our original problem asks to show that $n>v_3(8^n+1)$ for $n\ge 4$, we can plug this result from the LTE into our inequality,

$$n>v_3(n)+2$$

At this point it should be pretty much down hill, but let's write $n=3^t m$ for $v_3(m)=0$ to make it clearer to look at.

$$3^tm>t+2$$

An exponential grows much faster than linear, so it's proven. The only contradictions to this inequality occur for when $n<4$.

Merosity
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    Ah so what I've observed was actually LTE doing its job. I've never heard of it untill now so thank you for giving it a name. – Vepir Apr 17 '20 at 10:51