# Diophantine equation power of 7 and 2

$$7^x = 2^y \cdot 3 + 1$$

Find all positive $$(x,y) \in \mathbb{N}^2$$

When I look at this equation $$\mod 3$$ or $$\mod 7$$ it does hold - but how can I continue from here?

I know that $$7^x -1$$ is even so I can write it as: $$2k$$

$$2k = 2^y \cdot 3$$

$$2$$ does not divide $$3$$ and the same backwards - so $$3 \mid k$$ thus $$k \in \{3, 6, 9 , \dots \}$$ (not including $$0$$ because then $$x=0$$ which is not allowed)

Also $$2 \mid k$$ thus $$k \in \{2, 4, 6, 8, \dots \}$$

But again, I am stuck with a dead end - I am not sure how to continue from here.. I would appreciate your help, thank you!

• $(2,4)$ is a solution by inspection. Commented Sep 8, 2020 at 13:38
• @AlexeyBurdin Also $(1,1)$ but how do I generalize this? Commented Sep 8, 2020 at 13:38
• Clearly, $(1,1)$ is a solution (and there are no other solutions with $y=1$). If $y\geq 2$, we have $7^{x}\equiv 1\pmod 4$. Hence, $x$ is even (why?) and $x=2x_1$. Now we can rewrite our equation as $(7^{x_1}-1)(7^{x_1}+1)=2^y\cdot 3$. Can you end now? Commented Sep 8, 2020 at 13:39
• @richrow Why is $x$ even? I mean - it does work, but if I put $1$ or $3$ etc - the remainder isn't $1$ - is there a way to prove it? Commented Sep 8, 2020 at 13:41
• Stack there is a fine answer by Michael, about three hours ago. Commented Sep 8, 2020 at 17:54

Let $$x>2$$ and $$y>4$$.

Rewrite our equation in the following form: $$49(7^{x-2}-1)=48(2^{y-4}-1),$$ which says that $$2^{y-4}-1$$ is divisible by $$49$$,

which says that $$y-4$$ is divisible by $$21,$$ which says $$2^{y-4}-1$$ is divisible by $$2^{21}-1=49\cdot127\cdot337,$$

which gives that $$7^{x-2}-1$$ is divisible by $$337$$,

which says $$x-2$$ is divisible by $$56$$ (thanks to dear Will Jagy).

and from here $$7^{x-2}-1$$ is divisible by $$7^{56}-1=2^6\cdot3\cdot5^2\cdot29\cdot113...,$$

which gives $$48(2^{y-4}-1)$$ is divisible by $$64$$, which is a contradiction.

Id est, our equation has no natural solutions for $$x>2$$ and $$y>4$$.

Can you end it now?

• Thank you, Michael. I hope the others notice this. I did put the url for this in my (home computer) list. Commented Sep 8, 2020 at 16:11
• good, bumped in the active queue, while any 10K can still see my list of links Commented Sep 9, 2020 at 13:52
• @Will Jagy I found a simpler solution and fixed my post. Commented Sep 9, 2020 at 17:05
• @Will Jagy Yes, you are right. I missed it. I fixed. Thank you! Commented Sep 9, 2020 at 18:29
• @Stack I undeleted. Commented Sep 19, 2020 at 20:09

There is a very good method for

$$a p^m = b q^n + c,$$ where all are positive integers and $$p,q$$ are prime

Exponential Diophantine equation $7^y + 2 = 3^x$

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. ME! 41, 31, 241, 17

Finding solutions to the diophantine equation $7^a=3^b+100$ 343 - 243 = 100

http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847

The diophantine equation $5\times 2^{x-4}=3^y-1$

Equation in integers $7^x-3^y=4$

Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

Hello, Sailor

There was a girl in high school, active in "forensics" which was combined debate and related competition among many schools. She had practiced a really excellent Hello, Sailor. At the time, about 1974...

Eric Idle wrote Hello Sailor, his first novel, in 1970

A book of the same title was mentioned by Idle and Cleese in the Monty Python's Flying Circus episode "Sex and Violence" during "The Wrestling Epilogue" sketch, in which a humanist philosophy professor, author of a novel entitled "Hello Sailor," debates an Anglican monsignor over the existence of God in an officiated wrestling match.

https://en.wikipedia.org/wiki/Hello,_sailor

should bump question in active queue ... appears MIchael's answer does appear first maybe for being accepted .. Seems appropriate ... compare active queue after deleting

• Why did you post it? I wanted to show, how it works here. :) Commented Sep 8, 2020 at 14:02
• @MichaelRozenberg You are always welcome to ;) Thanks for the posts I will dig into them now! Commented Sep 8, 2020 at 14:04

I edit my previous answer. My only purpose here is to give an answer distinct from that given by the distinguished friend Michael Rozenberg.

We easily verify that $$y=1$$ and $$y=4$$ give two solutions and that $$y=2$$ and $$y=3$$ must be discarded; also $$x$$ must be even (reducing modulo $$16$$) so we consider the new equation $$7^{2x}=3\cdot2^{4+y}+1\iff(49)^x=48\cdot2^y+1;\space x\ge1, \space y\ge1$$ Now if $$x$$ is even then $$1\equiv8\cdot2^y+1\pmod{10}\Rightarrow 0\equiv2^{y+3}\pmod{10}$$ which is not possible so $$x$$ should be odd.

On the other hand we have $$(48+1)^x=48^2M+48x+1=48\cdot2^y+1\Rightarrow48M+x=2^y$$ and $$x$$ should be even.

Since $$x$$ cannot be odd and even,the only solutions of the proposed equation are $$(x,y)=(1,1),(2,4)$$

• Are you saying $(48+1)^x-1=\sum\limits_{k=0}^{x-1}48^{x-k}$ and that's odd?! Commented Sep 9, 2020 at 20:42
• I see now your comment. Thank you.What I was said is $$\sum_{k=0}^{x-1}48^{x-k}=48\cdot2^y$$ after eliminating $1$ in both sides. Commented Sep 9, 2020 at 22:56