Let $x < 81$, $x \equiv 1\pmod{3}$ and $x^3+x+1\equiv 0\pmod{81}$ find x

Question

First I used that $x \equiv 1\pmod{3}$, so that $x=3k+1$ for some $k$.

Then I put it in the next equation so $(3k+1)^3+(3+k)+1\equiv 81k^3+81k^2+12k+3\equiv 0\pmod{81}$,

So $12k+3\equiv 0\pmod{81}$ Then I divided it by $3$ (Is this possible and why is this possible?) to get $4k+1\equiv 0\pmod{27}$.

Next step was to find the inverse of 4 in $\mathbb{Z}/27\mathbb{Z}$ via Euclidean algorithm which is 7. So $k\equiv -7\equiv 20\pmod{27}$.

So I come to the solution that $x=20*3+1=61$ which works. However, I want to find a more elegant way or is this the normal approach to find a solution?

Your method is, as far as I know, the most normal method in this case. — S.C.B., Aug 01 '16 at 13:50
Look up Hensel's Lemma and Newton iteration to learn how it works in general. — Bill Dubuque, Aug 01 '16 at 13:51
Looks good to me. This generalizes to so called Hensel lifting, where under certain conditions you can "uniquely lift" a solution modulo a prime (here $x=1$ modulo $3$) to a solution modulo any power of the same prime. — Jyrki Lahtonen, Aug 01 '16 at 13:51
@Bernard See e,g. here. Btw, you need to explicitly ping users using at-sign if you want them to be notified in their inbox (I saw your comment only by coincidence when I revisited the question). — Bill Dubuque, Aug 01 '16 at 15:55

score 0 · Accepted Answer · answered Aug 01 '16 at 13:57

You may indeed divide by $3$ in $12k+3 \equiv 0 \pmod{81}$. Indeed, the solutions are precisely those $k$ such that there is $n$ with $12k+3 = 81n$; but that is precisely those $k$ such that there is $n$ with $4k+1 = 27n$; and that is precisely the statement that $4k+1 \equiv 0 \pmod{27}$.

As others have mentioned, Hensel lifting is a way to generalise this.

Let $x < 81$, $x \equiv 1\pmod{3}$ and $x^3+x+1\equiv 0\pmod{81}$ find x

1 Answers1