Solve over the positive integers: $7^x+18=19^y.$

Question

Solve over the positive integers:
$$7^x+18=19^y.$$

Progress:-
I first took $\mod 7,$ so we get $5^y\equiv 4 \mod 7$ since $5$ is a primitive root of $7$ and $5^2\equiv 4\mod 7.$ So we get $y\equiv 2\mod 6.$

And then took $\mod 9$
So we get $7^x\equiv 1\mod 9.$ Since residues of $7^x$ are $\{7,9,1\}.$
We get $x\equiv 0\mod 3.$

Then I couldn't get any progress, I tried Zsigmondy,etc stuff and also noticed $7^x-1=19(19^{y-1}-1)$

Any hints? Thanks in advance.

Answer 1

we suspect that $343 + 18 = 361$ is the largest solution. Assume we have a larger solution, I write that as $$ 7^3 (7^x-1) = 19^2 (19^y - 1) $$ with assumed $x,y \geq 1.$ Note that these are shifted from the $x,y$ values in the question.

String of calculations with simple conclusions about $x,y$

$ 19 | 7^x-1$ so $3|x$

$ 7 | 19^y - 1$ so $6|y$

calculate $8 | 19^6-1,$ so that $8 | 7^x - 1$

$8 | 7^x - 1,$ so that $2|x,$ cumulative $6|x$

calculate $43| 7^6 - 1 ,$ so that $43|19^y - 1$

$43|19^y - 1,$ so that $42|y$

calculate $7^4 | 19^{42} - 1$

However, with $x,y > 0,$ this tells us that $$ 7^4 | 7^3 (7^x-1) $$ As $7^x-1 \neq 0$ we see that $7^x-1$ is not divisible by $7,$ and so $ 7^4 | 7^3 (7^x-1) $ is a CONTRADICTION

next day: I was asked about the business with $19^y \pmod {43}.$ Notice how $19^{21} \equiv 42 \equiv -1 \pmod{43}, $ a square root of $1.$ Next, $19^{14} \equiv 36 \pmod{43} $ and $19^{28} \equiv 6 \pmod{43} ,$ while $6^3 = 216 \equiv 1 \pmod{43}, $ giving $6^3 \equiv 36^3 \equiv 1 \pmod{43} $

Mon May 10 10:16:00 PDT 2021
1   19
2   17
3   22
4   31
5   30
6   11
7   37
8   15
9   27
10   40
11   29
12   35
13   20
14   36
15   39
16   10
17   18
18   41
19   5
20   9
21   42
22   24
23   26
24   21
25   12
26   13
27   32
28   6
29   28
30   16
31   3
32   14
33   8
34   23
35   7
36   4
37   33
38   25
39   2
40   38
41   34
42   1    ***********
43   19
44   17
45   22
Mon May 10 10:16:00 PDT 2021