Prove that $2\cdot 3^x +1= p^y$ has no solution

Question

Prove that the Diophantine equation of $3$ variables $(x,y,p)$ $$2\cdot 3^x +1= p^y$$ has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.

I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradiction. But I cannot prove there is no solution for the case $y = 2k+1$.

I try to apply the technique in this answer (equation $7^x = 3 \cdot 2^y +1$) but it doesn't work as $p$ is not known.

Any help would be greatly appreciated!

Answer 1

Note that

$$2\cdot 3^x = p^y - 1 = (p - 1)(p^{y-1} + p^{y-2} + \ldots + p + 1) \tag{1}\label{eq1A}$$

With $p$ being odd and $p \gt 3$, then

$$p - 1 = 2(3^{a}) \; \; \to \; \; p = 2(3^{a}) + 1 \tag{2}\label{eq2A}$$

for some integer $a \ge 1$. Next, using the lifting-the-exponent (LTE) lemma, we have

$$x = \nu_3(p^y - 1) = \nu_3(p - 1) + \nu_3(y) = a + \nu_3(y) \tag{3}\label{eq3A}$$

Letting $\nu_3(y) = b$, there's then an integer $c \ge 1$ where $3 \nmid c$ and

$$y = c(3^b) \tag{4}\label{eq4A}$$

We next have

$$\begin{equation}\begin{aligned} p^y & = 2^y(3^{ay}) + y(2(3^{a}))^{y-1} + \ldots + y(2(3^{a})) + 1 \\ p^y - 1 & \gt 2^y(3^{ay}) \\ 2(3^x) & \gt 2^y(3^{ay}) \\ 3^{x} & \gt 3^{ay} \\ 3^{a+b} & \gt 3^{ac(3^b)} \\ a + b & \gt a(3^b) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

If $b = 0$, we get $a \gt a$, which is not true, so $b \gt 0$. Next, we have from \eqref{eq5A} that

$$a + b \gt a(3^b) \; \; \to \; \; b \gt a(3^b - 1) \ge 3^b - 1 \tag{6}\label{eq6A}$$

which is not possible. Thus, there's no solution to \eqref{eq1A} with $x \ge 2$, $y \ge 2$ and $p$ being a prime number (actually, no integer $p \gt 3$ works).

Answer 2

Here is a solution using Zsigmondy's theorem, perhaps less elementary than required in this case, but I think it is worth to include as it can be useful in a wide variety of similar problems.

Recall the theorem tells us that for coprime integers $a>b>0$ and any integer $n\geq 2$, we must have a prime $q\mid a^n-b^n$ such that $q\nmid a^k-b^k$ for all $k=1,2,\dots n-1$, except for a few special cases: when $n=2$ with $a+b$ being a power of $2$ or $n=6$ with $a=2,b=1$.

In this problem we have $$ 2\cdot 3^x=p^y-1=(p-1)(p^{y-1}+\dots+p+1)\tag{*} $$ and by Zsigmondy's theorem there is a prime $q\mid p^y-1$ with $q\nmid p-1$ (exceptions $y=2$ and $y=6$ are both even and were already ruled out by your argument). We cannot have $q=2$ as clearly $p$ is odd and so $2\mid p-1$. So $q=3$, and $3\nmid p-1$. Then by factorization in $(*)$ we must have $p-1=2$ and $p^{y-1}+\dots+p+1=3^x$. So $p=3$ and for $x\geq 1$ we get a contradiction $1\equiv 0\pmod 3$.

Answer 3

We have, as the O.P. noted, the exponent of the prime $p$ is odd so $$2\cdot3^x=(p-1)(p^{2y}+p^{2y-1}+\cdots+p+1)$$ the second factor having $y+1$ terms is odd so $2$ divides exactly $p-1=2m$, with $m$ odd. It follows $$3^x=m(p^{2y}+p^{2y-1}+\cdots+p+1)\Rightarrow m=3^a \text { and } \boxed{p=2\cdot 3^a+1} $$ then $$3^{x-a}=(2\cdot 3^a+1)^{2y}+(2\cdot 3^a+1)^{2y-1}+\cdots+(2\cdot 3^a+1)+1$$ which implies $$3^{x-a}\equiv {1+1+\cdots+1}=2y+1\pmod{3^a}\iff3^{x-a}=2y+1+3^aX$$ then $2y+1=3^ak$.

Back to the proposed equation we have $$2\cdot3^x+1=(2\cdot3^a+1)^{3^ak}$$ and similarly as above $$2\cdot3^x=2\cdot3^a[(2\cdot3^a)^{3^ak-1}+\cdots+(2\cdot3^a)+1]\Rightarrow 3^{x-a}\equiv 1\pmod{3^a}\iff3^{x-a}=1+3^aY$$ Thus $$3^a(3^{x-2a}-Y)=1\Rightarrow 3^a=1\Rightarrow a=0\Rightarrow p=3$$ We are done.