3

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

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!

CSch of x
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    $(2,4)$ is a solution by inspection. – Alexey Burdin Sep 08 '20 at 13:38
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    @AlexeyBurdin Also $(1,1)$ but how do I generalize this? – CSch of x Sep 08 '20 at 13:38
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    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? – richrow Sep 08 '20 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? – CSch of x Sep 08 '20 at 13:41
  • If $x=2x_1+1$ where $x_1\geq 0$, then $7^{x}=7^{2x_1+1}\equiv (-1)^{2x_1+1}=-1\pmod 4$. – richrow Sep 08 '20 at 13:43
  • @richrow Thank you! I tried to continue where you left off at $(7^k -1)(7^k +1) = 2^y \cdot 3$ (I called $x_1$ = $k$ just to make it easier to write) so I know that both of these terms are even ($7^k$ is odd $-1$ or $+1$ and it's even) but I got to a dead end here.. – CSch of x Sep 08 '20 at 13:46
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    Stack there is a fine answer by Michael, about three hours ago. – Will Jagy Sep 08 '20 at 17:54

3 Answers3

4

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?

Michael Rozenberg
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    Thank you, Michael. I hope the others notice this. I did put the url for this in my (home computer) list. – Will Jagy Sep 08 '20 at 16:11
  • good, bumped in the active queue, while any 10K can still see my list of links – Will Jagy Sep 09 '20 at 13:52
  • @Will Jagy I found a simpler solution and fixed my post. – Michael Rozenberg Sep 09 '20 at 17:05
  • Oh, very good. I wrote specific programs in C++ to search for brief (and with small primes, if possible) as command line. \ jagy@phobeusjunior \ ./prime_power_minus_one 7 168 \ 
    7^168 - 1    = 2^6 3^2 5^2 13 19 29 43 73 113 181 193 337 409 911 1201 2129 3361 4733 51031 309079 517553  cdot mbox{BIG} \\ 

7^168 + 1    = 2 17 449 673 121633 169553  cdot mbox{BIG} \\  jagy@phobeusjunior \\ ./order 7 337 \\

    337 56 = 2^3 * 7 \ parisize = 4000000, primelimit = 500000 \ ? n = 7^56 - 1 \ %1 = 211587613802425391637729361787678676290060193600 \ ? n % 337 \ %2 = 0

     – Will Jagy Sep 09 '20 at 17:50
  • It probably won't change anything much. , but the 337 step says $x-2$ is divisible by 56 \ prime_power_minus_one 7 56 \ 
    7^56 - 1    = 2^6 3 5^2 29 113 337 911 1201 2129 4733 517553  cdot mbox{BIG}  \\

7^56 + 1    = 2 17 449 673 169553  cdot mbox{BIG}   \\    YES, you still get your 64 = 2^6  The first one is right as is,  \\ order 2 49  \\

    49 21 = 3 * 7

     – Will Jagy Sep 09 '20 at 17:54
  • pari-gp \ ? n = 7^56 - 1 \ %1 = 211587613802425391637729361787678676290060193600 \ ? factor(n) \

    [ 2 6] \

    [ 3 1] \

    [ 5 2] \

    [ 29 1] \

    [ 113 1] \

    [ 337 1] \

    [ 911 1] \

    [ 1201 1] \

    [ 2129 1] \

    [ 4733 1] \

    [ 517553 1] \

    [ 515717329 1] \

    [13564461457 1] \

    ?

     – Will Jagy Sep 09 '20 at 18:16
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    @Will Jagy Yes, you are right. I missed it. I fixed. Thank you! – Michael Rozenberg Sep 09 '20 at 18:29
  • @WillJagy Hey Will, I remember that you posted a CW answer with a lot of links for solving diophantine equation similar to this one - however now it is gone, can you please recover it? It was very helpful! Thanks! – CSch of x Sep 18 '20 at 20:32
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    @Stack I undeleted. – Will Jagy Sep 19 '20 at 20:09
2

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

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

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.

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

Will Jagy
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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 $(x,y)=(1,1),(2,4)$

Piquito
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