How to solve $x^5\equiv 3 \pmod{19}$?

Question

How do I solve this equation: $$x^5 \equiv 3\pmod{19} $$ where all I know is Fermat's Little Theorem and operations on multiplicative group $\mathbb Z_p^*$?

Answer 1

Given $x^5=3$, we can seek a solution to $5k\equiv 1\bmod 18$ and then, using Fermat's Little Theorem, we will have

$x\equiv x^{5k}\equiv 3^k\bmod 19$

So how to solve $5k\equiv 1\bmod 18$? Multiplying by $4$ leads to $20k\equiv 2k\equiv 4$ and thus $k\in\{2,11\}$. Clearly $k\equiv 11\bmod 18$ is the correct solution, the other one having become an extraneous root because we used a multiplier that divides zero $\bmod 18$.

So $x\equiv 3^{11}\bmod 19$ which can be figured in either of two ways. First, by the standard squaring and multiplication approach:

$3^2\equiv 9, 3^4\equiv 9^2\equiv 5, 3^5\equiv 3×5\equiv15, 3^{10}\equiv 15^2\equiv 16, \color{blue}{3^{11}\equiv 3×16\equiv10}$.

Alternatively, if we recognize that $3$ is not a quadratic residue $\bmod 19$, we can render

$3^{11}\equiv 3^2×3^9\equiv 9×(-1)\color{blue}{\equiv 10}$.

There is a nice way to check this result from the repeating decimal block for $1/19$. To wit:

$1/19=0.\overline{052631578947368421}$.

Shifting the decimal point five places to the right should give the repeating block for $10^5/19$, which begins with $15$ and thus is clearly also the repeating block for $3/19$.

Answer 2

First note that

• $19$ is a prime number.
• $19$ does not divide $x$ (why?).

If we take the $4$th power , we obtain

$$ x^{20} \equiv 3^4 = 81 \equiv 5 \pmod{19} $$

From the fermat little theorem we get $$ x^{20} = x \cdot x^{19} \equiv x^2 \equiv 5\pmod{19} $$ So we are looking for $x$ such that $$ x^2 = 19\cdot n + 5 $$ If we try the first numbers we discover that for $n=5$ the RHS becames $100$, hence $x=10$ solves the equation.

Answer 3

From the above one as $x^2\equiv 5\pmod {19}$ we have $x^2\equiv 81\pmod {19}$ and therefore, the solutions are of the form $19k- 9$.

and for $x=19k+9$ is not the solution because, when we plug this into the original modulo, with the fact that $19\not|(9^5-3)$, the equivalence is not possible.