Show that 7 is a primitive root and use this to solve $7x^6 \equiv 6 \pmod {13}$.
Also find solutions to the non-linear Diophantine equation $7x^6 = 6 + 13y$.
The number 7 is a primitive root modulo 13 if the order of 7 modulo 13 is $\phi(13)=12$. Euler's Theorem tells us $7^{\phi(13)} \equiv 7^{12} \equiv 1 \pmod {13}$, and so we only need to test factors of 12, which are 1, 2, 3, 4, 6, 12.
| n | 7^n | 7^n mod 13 |
| 1 | 7 | 7 |
| 2 | 49 | 10 |
| 3 | 343 | 5 |
| 4 | 2401 | 9 |
| 6 | 117649 | 12 |
| 12 | 13841287201 | 1 |
The above calculations show that the order of 7 modulo 13 is indeed $\phi(13)=12$, and so 7 is a primitive root modulo 13.
Before proceeding, we need to prepare a table of indices.
| ind_7(a) = n | 7^n | a = 7^n mod 13 |
| 1 | 7 | 7 |
| 2 | 49 | 10 |
| 3 | 343 | 5 |
| 4 | 2401 | 9 |
| 5 | 16807 | 11 |
| 6 | 117649 | 12 |
| 7 | 823543 | 6 |
| 8 | 5764801 | 3 |
| 9 | 40353607 | 8 |
| 10 | 282475249 | 4 |
| 11 | 1977326743 | 2 |
| 12 | 13841287201 | 1 |
Applying Propositions (6.15) and (6.16) to $7x^6 \equiv 6 \pmod {13}$ gives
$$ \text{ind}_7(7) + 6 \times \text{ind}_7 (x) \equiv \text{ind}_7 (6) \pmod {12} $$
Using the table of indices gives us
$$ 6 \times \text{ind}_7 (x) \equiv 6 \pmod {12} $$
This has solutions if $g=\gcd(12,6)=6$ divides 6, which it does. In fact there are $g=6$ incongruent solutions.
Dividing through by 6
$$ \text{ind}_7 (x) \equiv 1 \pmod {2} $$
The 6 incongruent solutions modulo 12 are $\text{ind}_7(x) = 1, 3, 5, 7, 9, 11$.
The table of indices gives us
$$ x \equiv 2, 5, 6, 7, 8, 11 \pmod {13} $$
To solve $7x^6 = 6 + 13y$, we know the form of $x$ and only need the form of $y$.
For $x = 2 + 13t$, for some integer $t$, $ y = \frac{7(2+13t)^6 - 6}{13} $.
For $x = 5 + 13t$, for some integer $t$, $ y = \frac{7(5+13t)^6 - 6}{13} $.
For $x = 6 + 13t$, for some integer $t$, $ y = \frac{7(6+13t)^6 - 6}{13} $.
For $x = 7 + 13t$, for some integer $t$, $ y = \frac{7(7+13t)^6 - 6}{13} $.
For $x = 8 + 13t$, for some integer $t$, $ y = \frac{7(8+13t)^6 - 6}{13} $.
For $x = 11 + 13t$, for some integer $t$, $ y = \frac{7(11+13t)^6 - 6}{13} $.
The author's solution only uses $t=0$, and these solutions are as follows
$$ (2, 34), \quad (5, 8413), \quad (6, 25122) $$
$$ (2, 63349), \quad (5, 141154), \quad (6, 953917) $$