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

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

• 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. May 26, 2021 at 10:11

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 $$m=\pm1$$

• [+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 :) May 29, 2021 at 11:39

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 $$(6,43), (-4,-13)$$

• There are also the cases $n-1=\pm1$ and $n-1=-5$, which gives $p=-13$.
– lhf
May 26, 2021 at 10:29
• but for $n=-4$ I obtained $\frac{n^3-1}{5}=-13$ which is not a prime. May 26, 2021 at 10:57
• $-13$ 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. May 26, 2021 at 11:01
• Taking $n=4$ yields $5\nmid n^3-1$. I believe you meant $(6,43)$ May 29, 2021 at 12:14