$\textbf{Problem:}$Let $p$ be a prime number of the form $9k + 1$. Show that there exists an integer n such that $p | n^3 - 3n + 1$.


My only idea was that $n^3-3n+1$ might form a complete residue modulo such primes. So, I tried it out for $19$ and it turned out to be wrong. After that I could not find anything useful so I read the solution. But I don't understand it. Any elaboration on the solution given in the provided link or any new solution both are appreciated. Thanks in advance


Hint: $ x^3 - 3x + 1 $ is the minimal polynomial of $ \zeta_9 + \zeta_9^{-1} $ over $ \mathbf Q $, and if $ p \equiv 1 \pmod{9} $ then there is an element of order $ 9 $ in $ (\mathbf Z/p \mathbf Z)^{\times} $.


Elementary solution:

Suppose $n=t-1$, putting in we get:


For $t=3m$ we get:


We can assume $m^3-m^2=k$ so $9k+1|A$

Then $n=3m-1$

For example:

$m=2$, $\rightarrow:$, $p=37$, $n=5$

$m=3$, $\rightarrow:$, $p=163$, $n=8$

$m=4$, $\rightarrow:$, $p=433$, $n=11$

Hence values of n make an arithmetic progression .

  • 1
    $\begingroup$ How do you "assume $m^3 - m^2 = k$"? E.g. if $k=2$, then $ 9k+1 = 19$ is prime. What is your corresponding $m$ or $n$? $\endgroup$ – Calvin Lin Jan 18 at 6:48
  • $\begingroup$ @CalvinLin, see edit. $\endgroup$ – sirous Jan 18 at 7:05
  • $\begingroup$ As I asked, for prime $ p = 19$, what's the corresponding $m$ or $n$? You have shown that it is true for some primes of the form $ p = 9k+1$, not for all primes of the form. $\endgroup$ – Calvin Lin Jan 18 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.