Let $n$ a positive integer , show that $2^{n+3} \mid 3^n +7$ has no solutions?

Question

Motivation behind the problem :

We always could compute the $v_2(a^n+b^n)$ under some conditions by LTE !

What about expressions like $3^n+7$ , that are non factorizable ?

What I did :

$(1)$ I have checked that this is true for $n\leq 1000$

$(2)$ I showed that $n$ will be even , Here's why :

Suppose that $2^{n+3} \mid 3^n +7 \Rightarrow 8 \mid 3^n +7 \Rightarrow 3^n \equiv 1 \pmod8$

As $3^2\equiv 1 \pmod8 \Rightarrow 2 \mid n $ , thus $n$ is even !

Furthermore, if we let $n=2k$ :

$\Rightarrow 2^{n+3} \mid 9^k +7 = 9^k - 1 +8$

$\Rightarrow 2^{2k} \mid \displaystyle\frac{9^k-1}{8} +1$

$\Rightarrow \displaystyle\sum_{i=0}^{k-1}9^i \equiv - 1 \pmod{4^k} $

$\Rightarrow \displaystyle\sum_{i=0}^{k-1}9^i \equiv - 1 \pmod{4} $ , since $k\ge1$

As $9\equiv 1 \pmod4 \Rightarrow \displaystyle\sum_{i=0}^{k-1}1^i \equiv - 1 \pmod{4} $

$\Rightarrow k \equiv -1 \pmod4$

Finally, I get stuck here, but I still didn't found any contradiction yet.

Thanks for your help!

Answer 1

I'm not sure whether there is a solution only relying on elementary techniques, so let me show a general method depending on [Yu's version of p-adic Baker][$1$] (which thus might be a bit like nuking a mosquito).

Let $v_2$ denote the $p$-adic valuation of a non-zero rational number $a/b$. That is, if $a = p2^n$ and $b = q2^m$ for odd numbers $p$ and $q$, then $v_2(a/b) = n - m$. Thus, $v_2(12/5) = 2$.

We want to find natural numbers $n$ such that $v_2(3^n + 7) \ge n+3$, which is equivalent to $v_2(3^n \cdot \frac{-1}{7} - 1) \ge n+3$ when dividing by $-7$. Then, applying the result from Yu on $\Lambda = 3^n \cdot (-7)^{-1} - 1$ gives that $$ v_2(3^n \cdot (-7)^{-1} - 1) \le 19(20\sqrt{m+1}\cdot d)^{2(m+1)} e_{\frak{p}}^{m-1} \frac{p^{f_{\frak{p}}}}{(f_{\frak{p}} \log(p))^2} \log(e^5 m d) h_1 h_2 \log(B). $$ In general, Yu looks at forms $\alpha_1^{b_1} \alpha_2^{b_2} \cdots \alpha_n^{b_n} - 1$. Thus, we have $m = 2$, $\alpha_1 = 3$, $b_1 = n$, $\alpha_2 = -7$ and $b_2 = -1$. As we look at the $2$-adic valuation, $p = 2$. All three of $d$, $e_{\frak{p}}$, $f_{\frak{p}}$ are quite hard to explain in elementary terms, but because all our $\alpha_i$ are just rational numbers, we can take all three equal to $1$.
We have that $h_i = \max(h(\alpha_i), \log(p))$, where $h$ (is the again technical) absolute logarithmic Weil height. Again, as $\alpha_1$ and $\alpha_2$ are integers, we can take $h(\alpha_i) = \log(|\alpha_i|)$. Thus, $h_1 = \max(h(3), \log(p)) = \max(\log(3), \log(2)) = \log(3)$ and $h_2 = \max(h(-7), \log(2)) = \log(|-7|)$. Lastly, $B = \max(|b_1|, |b_2|) = \max(|n|, |-1|) = \max(n, 1)$. Filling this all in, we get that for $n \ge 1$, $$ v_2(3^n \cdot (-7)^{-1} - 1) \le 6.2 \cdot 10^8 \log(n) $$ (where I rounded the number up for convenience). Thus, for a solution to your inequality, we have to have $6.2 \cdot 10^8 \log(n) \ge n + 3$, which implies that $n \le 1.5 \cdot 10^{10}$.

Now we have a (quite inpractical) upper bound which can be reduced using the method in Vepir's answer in $3^n$ does not divide $4^n+5$ for $n\geq 2$.

[$1$]: Yu, K.: p-adic logarithmic forms and group varieties. II. Acta Arith. $89(4)$, $337–378$ $(1999)$, https://bibliotekanauki.pl/articles/1390501.pdf

Answer 2

The question already has an accepted answer by Derfellios, where is shown that $n\le 1.5\cdot 10^{10}$. But a new bounty was started, so I updated my answer and (provided I made no errors) showed that there is no such $n$.

It is easy to see that for each natural $k$, $3^{2^k}+1$ is divisible by $2$ and not divisible by $4$. Since $\left(3^{2^k}+1\right)\left(3^{2^k}-1\right)$, by induction on $k$ we can show that for each natural $k$, $3^{2^k}-1$ is divisible by $2^{k+2}$ and not divisible by $2^{k+3}$. Using this observation we can consecutively calculate the digits of the binary representation $\sum_{i=0}^\infty n_i\cdot 2^i$ of $n$:

$2^3|3^{0}+7$, so $n_0=0$.

$2^4|3^{2^1}+7$, so $n_1=1$.

$2^5|3^{2^1+2^2}+7$, so $n_2=1$.

$2^7|3^{2^1+2^2+2^3}+7$, so $n_3=1$ and $n_4=0$.

$2^9|3^{2^1+2^2+2^3+2^5}+7$, so $n_5=1$ and $n_6=0$.

$2^{12}|3^{2^1+2^2+2^3+2^5+2^7}+7$, so $n_7=1$, and $n_8=n_9=0$

$2^{20}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}}+7$, so $n_{10}=1$ and $n_{11}=\dots=n_{17}=0$.

$2^{21}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}+2^{18}}+7$, so $n_{18}=1$.

$2^{24}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}+2^{18}+2^{19}}+7$, so $n_{19}=1$ and $n_{20}=n_{21}=0$.

$2^{26}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}+2^{18}+2^{19}+2^{22}}+7$, so $n_{22}=1$ and $n_{23}=0$.

$2^{27}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}+2^{18}+2^{19}+2^{22}+2^{24}}+7$, so $n_{24}=1$.

$2^{38}|3^{2^1+2^2+2^3+2^5+2^7+2^{10}+2^{18}+2^{19}+2^{22}+2^{24}+2^{25}}+7$, so $n_{25}=1$ and $n_{26}=\dots=n_{35}=0$.

Since the last power of $3$ is not divisible by $2^{39}$, $n>2^{36}>1.5\cdot 10^{10}$.

Answer 3

Note that this solution is incomplete: I highlighted in bold the problem, that I don't see how to solve.

We will use the result, that follows from LTE lemma: $\nu_{2}(3^{2^{t}}-1) = t+2,\;t\geq 1$. Suppose that there is a solution. We would like to prove by induction, that $n$ is a sum of consecutive powers of two modulo power of two, i. e. $$n = 2^{l}k + \sum\limits_{i=1}^{l-1}2^{i},\;\{l,k\}\subset\mathbb{N}$$ Some of the base cases are covered by you: $n = 8k+6 = 8k + 4 + 2$, but we will need more. Let's prove that $n = 16k+14$. BWOC suppose that $n = 16k+6$. Thus, $$3^{n}+7 = 3^{6}(3^{16k}-1) + 3^{6}+7$$ Since $3^{16}-1\mid 3^{16k} - 1$ we have $\nu_{2}(3^{16k}-1)\geq \nu_{2}(3^{16}-1) = 6$. But $\nu_{2}(3^{6} + 7) = 5 < 6$, thus, $\nu_{2}(3^{n}+7) = 5 < n+3$. We would like to play the same game with $l=5$, but $\nu_{2}(3^{32}-1) = 7 = \nu_{2}(3^{14}+7)$. How do we deal with that? I will adress this case in the end.

Next, suppose that the claim is true for some $l\geq 5$ and by way of contradiction suppose that $$n = 2^{l+1}k + \underbrace{\sum\limits_{i=1}^{l-1}2^{i}}_{s}$$ Then $$3^{n}+7 = 3^{s}(3^{2^{l+1}k}-1) + 3^{s} + 7$$ Since $3^{2^{l+1}}-1\mid 3^{2^{l+1}k} - 1$ and $\nu_{2}(3^{2^{l+1}}-1) = l+3$, then $\nu_{2}(3^{2^{l+1}k} - 1) = l+3$. We now would like to prove that $\nu_{2}(3^{s}+7) < l+3$ But $$s = 2^{l}-2,$$ thus $\nu_{2}(3^{s}+7) = \nu_{2}(3^{2^{l}}+63) = \nu_{2}(3^{2^{l}}-1 + 64)$, and since $\nu_{2}(3^{2^{l}}-1) = l+2 > 6$, thus, $\nu_{2}(3^{s}+7) = \nu_{2}(3^{2^{l}}-1 + 64) = 6 < l+3$. Hence, $\nu_{2}(3^{n}+7) = 6 < n+3$ for $n\geq 4$.

So, $n$ is a sum of powers of two, thus, $n = 2^{t} - 2$ for some $t$. This will lead to the same contradiction, that we have just obtained.

In all cases above we were able to bound the exponent of $2$ in $3^{n}+7$ by a constant for some infinite set of integers. However, there is a particular problematic set, for which there is no such constant. This set consists of numbers $\{p_{i}\}_{i\in\mathbb{N}}$, which can be defined recursively: $$p_{1} = 14,\;p_{n+1} = 2^{\nu_{2}(3^{p_{n}}+7)-2} + p_{n}$$ Note that $$3^{p_{n+1}} + 7 = 3^{p_{n}}\left(3^{2^{\nu_{2}(3^{p_{n}}+7)-2}} - 1\right) + 3^{p_{n}} + 7 = 2^{\nu_{2}(3^{p_{n}}+7)}(3^{p_{n}}a + b),\;a\equiv b\equiv 1\mod 2,$$ so $$\nu_{2}(3^{p_{n+1}} + 7) > \nu_{2}(3^{p_{n}}+7)$$

Exponents of two that we add recursively form the following set: $$\{2^{5},2^{7},2^{10},2^{18},2^{19},2^{22},2^{25},2^{36},\ldots\}$$ It seems that they grow not so fast, but for now I don't know if there is an elementary way to come up with a bound, even a crude one.

Answer 4

Just so you know, the exponent for $2$ becomes quite small compared with the exponent for $3.$ I had it print out only when the $2$ exponent was larger than all previous $2$ exponents. The first line says that $10 = 2 \cdot 5.$ The second line says $9+7 = 16 = 2^4.$ And so on

 3 exponent:  1    2 exponent:  1 
 3 exponent:  2    2 exponent:  4 
 3 exponent:  6    2 exponent:  5 
 3 exponent:  14    2 exponent:  7 
 3 exponent:  46    2 exponent:  9 
 3 exponent:  174    2 exponent:  12 
 3 exponent:  1198    2 exponent:  20 
 3 exponent:  263342    2 exponent:  21 
 3 exponent:  787630    2 exponent:  24 

Answer 5

One result I've found , may be it's interesting !

$3^n + 7 = 2^{\alpha} \Rightarrow n=2 $ and $\alpha = 4$

Proof :

$3^n + 7 = 2^{\alpha} \Rightarrow 3^n +6 = 2^{\alpha} - 1 \Rightarrow 3 \mid 2^{\alpha} -1$

$\Rightarrow 2^{\alpha} \equiv 1 \pmod 3$ , As $2^2 \equiv 1 \pmod 3 \Rightarrow 2 \mid \alpha$ , so $\alpha$ is even !

$3^n + 7 = 2^{\alpha} \Rightarrow 3^n +1 = 2^{\alpha}-6 = 2^1 \times ( 2^{\alpha-1}-3) \Rightarrow v_2(3^n+1) = 1 \Rightarrow n$ is also even !

Using $\alpha=2\gamma$ and $n=2\beta$ , we get the following :

$9^{\beta}+6=4^{\gamma}-1 \Rightarrow 3^{2\beta-1}+2 =\displaystyle\frac{4^{\gamma}-1}{3} = \displaystyle\sum_{i=0}^{\gamma-1}4^i$

As $4\equiv 1 \pmod 3 \Rightarrow \displaystyle\sum_{i=0}^{\gamma-1}1^i \equiv 2 \pmod3 \Rightarrow \gamma \equiv 2 \pmod3$

Now, we take $\gamma = 2 + 3y \Rightarrow 9^{\beta}+7=4^{2+3y}$

As $4^3 \equiv 1 \pmod 7 $ and $4^2 \equiv 2 \pmod 7 \Rightarrow 4^{2+3y} \equiv 2 \pmod 7 $

$\Rightarrow 9^{\beta} \equiv 2 \equiv 9 \pmod 7 \Rightarrow 9^{\beta-1} \equiv 1 \pmod 7$

As $9^3 \equiv 1 \pmod 7 \Rightarrow 3\mid \beta - 1 \Rightarrow \beta \equiv 1 \pmod 3 $

Finally, we use both $\gamma = 2 + 3y$ and $\beta = 1 + 3x$ :

$\Rightarrow n = 2 + 6x$ and $\alpha = 4 +6y$

$\Rightarrow (2^{3y+2})^2 - (3^{3x+1})^2 = 7 $

$\Rightarrow (2^{3y+2} - 3^{3x+1})(2^{3y+2} + 3^{3x+1}) = 7 $

$\Rightarrow \begin{cases} 2^{3y+2} - 3^{3x+1} = 1 \\\\ 2^{3y+2} + 3^{3x+1} = 7\end{cases} \Rightarrow 2 \times 2^{3y+2} = 8 \Rightarrow 2^{3y+3} = 2^3$

$\Rightarrow 3y+3=3 \Rightarrow y=0 $ and similarly, we get $x=0$

Thus $n=2 $ and $\alpha = 4$