how to solve to congruence $x^{98}\equiv99\pmod{125}

Question

I show you my attempt: $$(125, 98) = 1 \Rightarrow(x^{98} , 125) = 1 \Rightarrow (x, 125) = 1$$ (Euclidian Algorithm)

$$x^{\phi(125)} = x^{100} \equiv1\pmod{125} \wedge x^{98} \equiv99(\mod 125) \Rightarrow x^2\equiv99 \pmod{125} \Rightarrow x^2 \equiv{99}\pmod{5} \Rightarrow x^2 \equiv{4}\pmod{5}$$

Thus, $$5|(x-2)(x+2)$$ $$x\equiv{7,3}\pmod{5}$$ $$x\equiv{7,3}\pmod{125}$$

I ask you to check my solution and explain me when I got wrong.

Answer 1

HINT:

$$x^{100}\equiv1,x^{98}\equiv99\implies x^2(x^{98})\equiv1\implies99x^2\equiv1\pmod{125}\ \ \ \ (1)$$

$(1)\implies99x^2\equiv1\pmod5\iff-x^2\equiv1\iff x^2\equiv-1\equiv4\implies x\equiv\pm2$

Now use Hensel's lemma (1), (2)

Answer 2

We have: $$x^2\equiv \frac{1}{99}\equiv 24\pmod{125}.\tag{1}$$ Since $8\cdot 125+24 =1024 = 32^2$ it follows that $x\equiv 32\pmod{125}$ is a solution of $(1)$.

Since $(\mathbb{Z}/125\mathbb{Z})^*$ is a cyclic group, the only other solution is $x\equiv -32\equiv 93\pmod{125}$.