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

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!

  • 1
    $\begingroup$ $(2,4)$ is a solution by inspection. $\endgroup$ Sep 8, 2020 at 13:38
  • 1
    $\begingroup$ @AlexeyBurdin Also $(1,1)$ but how do I generalize this? $\endgroup$
    – CSch of x
    Sep 8, 2020 at 13:38
  • 2
    $\begingroup$ 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? $\endgroup$
    – richrow
    Sep 8, 2020 at 13:39
  • $\begingroup$ @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? $\endgroup$
    – CSch of x
    Sep 8, 2020 at 13:41
  • 1
    $\begingroup$ Stack there is a fine answer by Michael, about three hours ago. $\endgroup$
    – Will Jagy
    Sep 8, 2020 at 17:54

3 Answers 3


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?

  • 1
    $\begingroup$ Thank you, Michael. I hope the others notice this. I did put the url for this in my (home computer) list. $\endgroup$
    – Will Jagy
    Sep 8, 2020 at 16:11
  • $\begingroup$ good, bumped in the active queue, while any 10K can still see my list of links $\endgroup$
    – Will Jagy
    Sep 9, 2020 at 13:52
  • $\begingroup$ @Will Jagy I found a simpler solution and fixed my post. $\endgroup$ Sep 9, 2020 at 17:05
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    $\begingroup$ @Will Jagy Yes, you are right. I missed it. I fixed. Thank you! $\endgroup$ Sep 9, 2020 at 18:29
  • 1
    $\begingroup$ @Stack I undeleted. $\endgroup$
    – Will Jagy
    Sep 19, 2020 at 20:09

CW answer, votes don't affect me for this one.

There is a very good method for

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

discovered by https://math.stackexchange.com/users/292972/gyumin-roh

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


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 : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. 128 - 125 = 3

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.


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

  • $\begingroup$ Why did you post it? I wanted to show, how it works here. :) $\endgroup$ Sep 8, 2020 at 14:02
  • $\begingroup$ @MichaelRozenberg You are always welcome to ;) Thanks for the posts I will dig into them now! $\endgroup$
    – CSch of x
    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 $(x,y)=(1,1),(2,4)$

  • $\begingroup$ Are you saying $(48+1)^x-1=\sum\limits_{k=0}^{x-1}48^{x-k}$ and that's odd?! $\endgroup$ Sep 9, 2020 at 20:42
  • $\begingroup$ 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. $\endgroup$
    – Piquito
    Sep 9, 2020 at 22:56

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