TITLE: Solve over the positive integers: $7^x+18=19^y.$
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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