A Conjecture about the upper bound of $v_2(3^n+d)$

Question

The Conjencture :

$\forall d \in \mathbb{N^*}, \exists n_0 \in \mathbb{N^*}$ such that : $\forall n>n_0 : v_2(3^n+d)<n$ , where $n$ is a positive integer.

$v_2(x)$ is the $2$-adic valuation of $x$

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

At the problem above, we need an upper bound of $v_2(3^n+d)$ at $d=7$

My attempt:

If $d$ is even, we are already done ! , since $v_2(3^n+d)=0<n$

Now let's take $d\in 2\mathbb{N}+1$ :

($1$) $d\equiv 1 \pmod8 \Rightarrow 3^n+d \equiv 3^n + 1\pmod8$

If $n$ is even $\Rightarrow 3^n+1 \equiv 2 \pmod8 \Rightarrow v_2(3^n+d)=1 <n$, $n_0=1$

If $n$ is odd $\Rightarrow 3^n+1 \equiv 4 \pmod8 \Rightarrow v_2(3^n+d)=2 <n$, $n_0=2$

($2$) $d\equiv 3 \pmod8 \Rightarrow 3^n+d \equiv 3^n + 3\pmod8$

If $n$ is even $\Rightarrow 3^n+3 \equiv 4 \pmod8 \Rightarrow v_2(3^n+d)=2 <n$, $n_0=2$

If $n$ is odd $\Rightarrow 3^n+3 \equiv 6 \pmod8 \Rightarrow v_2(3^n+d)=1 <n$, $n_0=1$

So, If $d\equiv 1 \pmod8$ or $d\equiv 3 \pmod8$ , the conjencture holds !

Thus, you can restrict to $d \equiv 5,7 \pmod 8$

Answer 1

Let me give a quick answer based on the answers in the comments from Torsten Schoeneberg, Mike Bennett and myself (the original question asker asked for this). I have copied most of my answer on the linked answer.

Following the answer in the linked question, and again apply Yu's version of p-adic Baker [$1$] on $3^n + d$.

We want to find natural numbers $n$ such that $v_2(3^n + d) \ge n$, which is equivalent to $v_2(3^n \cdot \frac{-1}{d} - 1) \ge n$ when dividing by $-d$. Then, applying the result from Yu on $\Lambda = 3^n \cdot (-7)^{-1} - 1$ gives that when $\Lambda \ne 0$ (which is equivalent to $3^n \ne d$), we have 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 $\alpha_1^{b_1} \alpha_2^{b_2} \cdots \alpha_n^{b_m} - 1$. Thus, we have $m = 2$, $\alpha_1 = 3$, $b_1 = n$, $\alpha_2 = -d$ 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, things are little easier, and for integers $n \ne 0$, $h(n) = \log|n|$. Thus, $h_1 = \max(h(3), \log(p)) = \max(\log(3), \log(2)) = \log(3)$ and $h_2 = \max(h(-d), \log(2)) = \log(d)$ for $d \ge 1$. As all these numbers are computable, we get that for some constant $C > 0$, that $$ n \le v_2(3^n \cdot (-7)^{-1} - 1) \le C \cdot \log(d) \cdot \log(n) \tag{$\star$} $$ when $3^n \ne d$. As $3^n = d$ and $n \le C \cdot \log(d) \cdot \log(n)$ both have finitely many solutions $n \in \mathbb{N}$, $v_2(3^n + d) \ge n$ has only finitely many solutions. Thus, the conjecture holds!

When $d \ge 2$, one option for $n_0$ is $(C \log(d))^2$, where $C$ is as computed above. Then, for any $n \ge n_0$, $n > \log_3(d)$ and so $3^n + d \ne 0$. Moreover, for $n \ge n_0$, $$\frac{n}{\log(n)} \ge \frac{n_0}{\log(n_0)} = C\log(d)\frac{C\log(d)}{2\log(C\log(d))} \ge \frac{n}{\log(n)} \frac{C\log(d)}{2\log(C\log(d))},$$ where the last inequality is due to $(\star)$. This is impossible as $\frac{C\log(d)}{2\log(C\log(d))} > 1$. Thus, our choice for $n_0$ is correct (but is of course not necessarily optimal).

Let me now briefly break down what this formula does, and what it has to do with $p$-adic logarithms. The $2$-adic logarithm (which I denote with $\log_2$) is defined on all integers that are $1$ modulo $4$. In particular, we have that when $x \equiv 1 \mod 4$, $v_2(\log_2(x)) = v_2(x-1)$. We can extend the notion of being $1$ modulo $4$ to the rationals by demanding that the numerator of the fraction is equivalent to $1$ modulo $4$. Taking $x = -3^n/d$ and assuming that $v_2(-3^n/d-1) = v_2(3^n + d) \ge n \ge 2$, we indeed have this property and so $v_2(3^n + d) = v_2(\log_2(-3^n/d))$.

If $d \equiv 1 \mod 4$, $n$ is odd and so setting $n = 2m+1$, we have that : $$2m+1 \ge v_2(\log_2(9^m \cdot -3/d)) = v_2(m\log_2(9) + \log_2(-3/d)) = \quad v_2(m + \log_2(-3/d)/\log_2(9))+3.$$ (Here, we used $v_2(\log(9)) = 3$.) Thus, we basically get a the following question: Is there an $m$ such that adding it to $\log_2(-3/d)/\log_2(9)$ has an enormous $2$-adic valuation? Or in other words, in the $2$-adic expansion of $\log_2(-3/d)/\log_2(9)$, are the $\approx m$ $1$'s in a row in the $2$-adic expansion of $\log_2(-3/d)/\log_2(9)$ when starting at index $\approx \log(m)$ ? Yu's methods show that this cannot happen for large $m$. Taking for example, $d = 5$, $$ \log_2(-3/d)/\log_2(9) = 1 + 2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^8 + 2^{13} + 2^{14} + 2^{15} + 2^{16} + O(2^{17}) $$ and so the smallest $m$ such that $v_2(m + \log_2(-3/d)/\log_2(9)) \ge 7$ is $m = 5$. This corresponds to $n = 2\cdot 5+1=11$, and then indeed $$ v_2(3^n+d) = 10 = v_2(m + \log_2(-3/d)/\log_2(9))+3. $$

Similarly, if $d \equiv 3 \mod 4$, $n$ is even and so setting $n = 2m$, we have : $$2m \ge v_2(\log_2(9^m \cdot -1/d)) = v_2(m\log_2(9) + \log_2(-1/d)) = \quad \quad v_2(m + \log_2(-1/d)/\log_2(9))+3,$$ giving almost the same situation.

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