4

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

Source. Here.

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

user26857
  • 52,094
Yes it's me
  • 1,854
  • 6
  • 15
  • 1
    How much do you know about rings/fields? Group theory? This will be required to explain the given answer. – Sarvesh Ravichandran Iyer Jan 18 '21 at 05:15
  • 1
    For that substitution and other details, you can read : https://math.stackexchange.com/questions/3814411/if-a-prime-natural-number-p-neq-3-divides-a3-3a1-for-some-integer-a-the?rq=1 and https://math.stackexchange.com/questions/3800911/prime-numbers-which-divide-n3-3n1 – Sarvesh Ravichandran Iyer Jan 18 '21 at 06:08

2 Answers2

5

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} $.

Ege Erdil
  • 17,747
0

Elementary solution:

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

$A=n^3-3n+1=t^3-3t^2+3$

For $t=3m$ we get:

$A=27m^3-27+3=3[9(m^3-m^2)+1]$

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 .

sirous
  • 10,751
  • 1
    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 $n$? – Calvin Lin Jan 18 '21 at 06:48
  • @CalvinLin, see edit. – sirous Jan 18 '21 at 07:05
  • 1
    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. – Calvin Lin Jan 18 '21 at 17:08