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Find all integers n such that $(\frac{n^3-1}{5})$ is prime?

My Approach:

I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.

$n^3-1=10,15,25,35,55,...,215$

which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$

My doubt is that how to check more value of $n$ without using modular arithmetic because the book I'm referring has not introduced it yet.

Second approach: $\frac{(n^3-1)}{5}=\frac{(n-1)(n^2+n+1)}{5}$

But my second approach too does not lead me anywhere.

This problem is from the book Pathfinder for Olympiad Mathematics

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  • $\begingroup$ The factorization in your second approach is the key. One of $n-1$ or $n^2+n+1$ has to be divisible by five. Basic congruence stuff tells you which it is. Call it $A$ and the other $B$. Then you have two possibilities. Either $B$ is prime and $A/5=1$. Or $A/5$ is a prime and $B=1$. The rest is elementary. $\endgroup$ May 26, 2021 at 10:11

2 Answers 2

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If we split all integer $n$ into $[5m+1,5m+2,5m+3,5m+4,5m+5]$ you can show that only numbers $(5m+1)^3-1$ are divisible by $5$ as $[2^3-1,3^3-1,4^3-1,5^3-1]$ are all not divisible by $5$

Now $$\frac{(5m+1)^3-1}{5}=3m+15m^2+25m^3=m(3+15m+25m^2)$$

and the product $m(3+15m+25m^2)$, can only be a candidate prime if $m=\pm1$

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  • $\begingroup$ [+1]! Technically, you should also check the case when $3+15m+25m^2=\pm 1$. But it is easy to notice that this equations won't yield a integer solution :) $\endgroup$
    – Dr. Mathva
    May 29, 2021 at 11:39
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First, since you need the fraction to be a prime, you have $n^3-1 = 5p$ with $p$ a prime number. Now, $n^3-1=(n-1)(n^2+n+1)$. So this leads to the idea that one of those factors is $5$ and the other $p$.

  • If $n-1=5$, $n=6$ so $n^2+n+1= 36+6+1= 43$.

  • If $n-1\neq 5$, $n^2+n+1=5$, so $n(n+1)=4$, and you can check in multiple ways that this equation doesn't have solutions. (one way is noticing the parity of the lhs and forcing the odd factor to be $1$, so the other won't reach to be $4$)

EDIT: As how mentioned @lhf , I'm missing if $n^3 - 1 = -5p$ So considering this case, you have these cases:

  • If $n-1 = -5$, $n = -4$, so $n^2+n+1 = 16 - 4 +1 = 13$

  • If $n-1 \neq -5$, then $n^2+n+1 = -5$, and this leaves to $n^2+n = -6$, obtaining with general formula this doesn't give any possible solution.

Also,

  • If $n-1 = -5p$, $n^2+n+1=1$, but this will mean $(5p)^2-5p=0$, implying $5p(5p-1)=0$, forcing $5p-1=0$, but that leaves $p$ not being a prime.

  • If $n-1 = 5p$, then $n = 5p + 1$, so $(5p+1)^2 + 5p + 1 +1 = 0$, giving $25p^2+15p + 3 = 0$, but this equation doesn't have any integer solution, since this would mean $5\mid 3$, which is a nonsense.

Then the solutions are $(n,p)$ are $(6,43), (-4,-13)$

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    $\begingroup$ There are also the cases $n-1=\pm1$ and $n-1=-5$, which gives $p=-13$. $\endgroup$
    – lhf
    May 26, 2021 at 10:29
  • $\begingroup$ but for $n=-4$ I obtained $\frac{n^3-1}{5}=-13$ which is not a prime. $\endgroup$
    – mathophile
    May 26, 2021 at 10:57
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    $\begingroup$ $-13$ is a prime. The definition I have of a prime number $p$ is a number that is divisible by $\pm 1$ and $\pm p$. If you're considering as a prime only the numbers such that are only divisible by $1$ and themselves, then you can consider $-13$ isn't a prime, but the definition I gave you is the one for primes. $\endgroup$
    – iam_agf
    May 26, 2021 at 11:01
  • $\begingroup$ Taking $n=4$ yields $5\nmid n^3-1$. I believe you meant $(6,43)$ $\endgroup$
    – Dr. Mathva
    May 29, 2021 at 12:14

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