Simple question about linear congruence with Fermat's little theorem $4x≡11\bmod19$

Question

$$4x≡11\bmod19$$ How to solve this with Fermat's theorem? I know it can be done with Euclidean algorithm and Bezout's coefficients as well. $$\gcd(4,19)=1$$ $$1=5 \cdot4-1\cdot19$$ $$\begin{align}4\cdot5x&≡5\cdot11\bmod19\\ &≡55\bmod19\\ &≡17\bmod19\end{align}$$ But with Fermat's theorem: $$4^{18}≡1\bmod19$$ How do I continue from here?

Answer 1

Hint: Multiply both sides of $4x\equiv 11\bmod 19$ by $4^{17}$.

Answer 2

Little Fermat's theorem just helps you finding $5$ directly, the catch is that it is cloaked under the form $4^{17}$.

But you can reduce it by dichotomy and various tricks

$4^{17}\equiv 4\times 4^{16}\equiv 4\times 16^8\equiv 4\times(-3)^8\equiv 4\times 9^4\equiv 4\times (-10)^4\equiv 40000\equiv 200^2\equiv (190+10)^2\equiv 100\equiv (19+1)\times 5\equiv 5\pmod {19}$

Answer 3

This doesn't seem to be a very good application of the Little Fermat Theorem, due to the amount of calculation required, as shown in zwim's answer above. Another approach would be:

$4x≡11\bmod19$ $\Longleftrightarrow 4x≡-8\bmod19 \Longleftrightarrow x≡-2\bmod19 \Longleftrightarrow x≡17\bmod19$