Solve the non-linear Diophantine equation
$$ x^6 = 1 + 13y $$
The solutions for $x$ are the solutions to the congruence
$$ x^6 \equiv 1 \pmod {13} $$
We know 2 is a primitive root of prime 13. The following is a table of indices.
| ind_2 (a) | a = 2^ind_r(a) mod 13 |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
| 4 | 3 |
| 5 | 6 |
| 6 | 12 |
| 7 | 11 |
| 8 | 9 |
| 9 | 5 |
| 10 | 10 |
| 11 | 7 |
| 12 | 1 |
Applying Propoositions (6.15) and (6.16) to $ x^6 \equiv 1 \pmod 13 $ gives
$$ 6 \times \text{ind}_2(x \equiv 0 \pmod {12} $$
Since $g=\gcd(12,6)=6$ divides 0, the linear congruence has solutions, and in fact gas $g=6$ incongruent solutions modulo 12.
Dividing by 6 gives
$$ \text{ind}_2(x) \equiv 0 \pmod 2 $$
This gives us $\text{ind}_2(x) \equiv 2, 4, 6, 8, 10, 12 \pmod {12}$. Using the table of indices above gives us
$$ x \equiv 1, 3, 4, 9, 10, 12 \pmod {13} $$
Using $y=\frac{x^6-1}{13}$ gives us the corresponding values for $y$, and so the solutions are
$$ (x,y) \quad = \quad (1,0), \; (3,56), \; (4,315), \; (9,40880), \; (10,76923), \; (12,229691) $$