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Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers. I couldn't have much progress.

Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.

Assume $y \ge 3$ and $x \ge 1$.

By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.

By $\mod 7$,we get $y \equiv 1 \mod 6$.

I also tried $\mod 2$ but it didn't work.

Please post hints ( not a solution). Thanks in advance!

J. W. Tanner
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Sunaina Pati
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  • Well, one easily eliminated case: if $x=2k$ were even then $7^{2k}-4=(7^k-2)(7^k+2)$ so... – lulu Nov 13 '20 at 01:11
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    General remark: since there is a solution, namely $(1,1)$, congruences alone won't get it done, though of course you can use congruences to eliminate a lot of cases. – lulu Nov 13 '20 at 01:13
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    @lulu there is a method that usually wors with small numbers http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7/1942409#1942409 and many others – Will Jagy Nov 13 '20 at 01:22
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    @SunainaPati FYI, your equation is $7^x - 3^y = 4$. For $x, y \gt 1$, the table in the Generalization section of Wikipedia's "Catalan's conjecture" article shows for a difference of $4$, there are only $3$ solutions with perfect powers less than $10^{18}$. These are $8 - 4 = 2^3 - 2^2$, $36 - 32 = 6^2 - 2^5$ and $125 - 121 = 5^3 - 11^2$. Thus, there's no solution for powers of $7$ and $3$ within that range, implying (but not proving) there are no other solutions than the $(1,1)$ one you've already found. – John Omielan Nov 13 '20 at 01:22
  • @WillJagy Ah, thank you. I have seen such things before, and was trying to reconstruct the method. Glad to have the reference. – lulu Nov 13 '20 at 01:23
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    @SunainaPata Also note later in that article, Pillai's conjecture states "... each positive integer occurs only finitely many times as a difference of perfect powers ...". – John Omielan Nov 13 '20 at 01:24
  • @lulu I have a file of url's at home, pasted a bunch of examples into a CW answer – Will Jagy Nov 13 '20 at 01:27
  • @JohnOmielan there is a successful method, see links in my CW answer – Will Jagy Nov 13 '20 at 01:28
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    @WillJagy Thanks for your feedback. I was just reading your answer for this specific equation when you commented. – John Omielan Nov 13 '20 at 01:29
  • @JohnOmielan good point, I was not sure I did this particular one. – Will Jagy Nov 13 '20 at 01:46

2 Answers2

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examples to study

CW:Catalan Thue Ramanujan Nagell Tijdeman p^x - q^y = C

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

Diophantine equation power of 7 and 2

Find natural numbers a,b such that $|3^a-2^b|=1$ did +-1

Will Jagy
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It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.

Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).

Therefore, $3^a-1$ is a multiple of $13$.

Therefore, $7^b-1$ is a multiple of $13$.

Therefore, $b$ is a multiple of (what?).

Therefore, $7^b-1$ is a multiple of $9$.

Therefore, $3(3^a-1)$ is a multiple of $9$.

Therefore, $a$ is (what?).

Therefore, $x$ is (what?).

Martin Sleziak
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J. W. Tanner
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