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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.

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  • $\begingroup$ $(3,2)$ is a solution as is $(0,1)$ so be warned that if you were hoping to prove no solutions exist via modular arguments that will fail. $\endgroup$
    – JMoravitz
    Commented May 9, 2021 at 18:36

2 Answers 2

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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
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  • $\begingroup$ Oh wow.. Really nice! $\endgroup$ Commented May 10, 2021 at 2:11
  • $\begingroup$ Hi. How do you get directly $$ 43|19^y - 1 \implies 42|y $$? $\endgroup$ Commented May 10, 2021 at 2:22
  • $\begingroup$ @lonestudent it is not automatic. In essence, calculate $19^t \pmod {43}$ for increasing $t$ until it becomes $1.$ We do know that the "order" is a positive divisor of $42$ $\endgroup$
    – Will Jagy
    Commented May 10, 2021 at 2:26
  • $\begingroup$ Isn't that discrete logarithm? $\endgroup$ Commented May 10, 2021 at 2:41
  • $\begingroup$ This is a discrete logarithm: alpertron.com.ar/DILOG.HTM $\endgroup$ Commented May 10, 2021 at 17:30
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Excluding the obvious solution $(x, y) = (0, 1)$, an argument mod $7$ shows that $y$ is even (which you have done). Similarly, an argument mod $19$ shows that $x$ is a multiple of $3$.

Thus we may look at the elliptic curve $Y^2 = X^3 + 18$. A computer algebra system such as Sage can be used to find all integral points: $(X, Y) = (7, \pm 19)$.

Therefore the only remaining solution is $(x, y) = (3, 2)$.


There might be elementary solutions, but this is the most efficient way of solving it.

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  • $\begingroup$ I am totally unfamiliar with computer algebra systems. Did Sage determine that it is impossible for there to be any other integral points besides the ones that you mentioned? If so, is it possible to read about the algorithm that Sage used to conclude that? $\endgroup$ Commented May 9, 2021 at 19:36
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    $\begingroup$ Yes, Sage can verify that there is no other integral points on this curve (excluding the infinite point). Here you can find documentation for the function integral_points and it is mentioned that the algorithm used is in Henri Cohen, Number Theory, Vol. I: Tools and Diophantine Equations. GTM 239, Springer, 2007. $\endgroup$
    – WhatsUp
    Commented May 9, 2021 at 20:19

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