Prove that $f(x)=x^{3} + 2x^{2}+ 3x + 5$ has a root in $\mathbb{Z}_{13}$ (The 13-adics).
EDIT: NOTE: I HAD A TYPO WITH THE POLYNOMIAL'S COEFFICIENT!!!.
QUESTION: Prove that $f(x)=x^{3} + 3x^{2}+ 3x + 5$ has a root in the in $\mathbb{Z}_{13}$. Then find the first two terms of the succession
My attempt: Using Hensel's lemma, we obtain that $f(2)$ has a solution mod 13 and is not congruent to 0 when evaluated in $f´(2)$. Thus it has a root in the 13-adics.
Now lets find the second term of the sequence $(2, a_1, a_2, ...)$. By Hensel's Lemma, $a_1 \equiv2$ mod 13. Then $a_1= 2+13t$., Then when evaluating $f(2+13t)$ and solving for t in Wolfram, we get that $t \equiv 5$ mod 13. thus $a_1 = 2+ 13(5) = 67$.
My question is, how do I construct the sequence that the problem requires?
Thank you in advance.
Suspecting that @OscarLanzi is correct about a mis-translation: let's test whether $x^3+3x^2+3x+4=0$ has a root in $\mathbb Z_{13}$. We first check in $\mathbb Z/13$, both for existence or not. As @Dan_Fulea observes (which is the kind of thing to hope for if there'd be a truly-human-understandable computation here), since $(x+1)^3=x^3+3x^2+3x+1$, the equation can be rearranged to $(x+1)^3+4=0$. Some small tricks to avoid computation: this is asking whether $-4$ is a cube mod $13$. Equivalently, is $4$ a cube? Since $4$ is a square, if it were also a cube, it'd be a sixth power (!). Since $(\mathbb Z/13)^\times$ is cyclic of order $13-1=12$, $-1$ is the only sixth power other than $1$, and $-1\not=4$ mod $13$. So there's no solution mod $13$...
EDIT: if there were a solution in the p-adic integers $\mathbb Z_{13}$, it would give a solution in $\mathbb Z_{13}/(13\cdot \mathbb Z_{13})\approx \mathbb Z/13\mathbb Z$. (Yes, this isomorphism needs a little proof...)