Find all primes $p$ such that $p^2-p+1$ is a perfect cube

Question

Find all primes $p$ such that $p^2-p+1$ is a perfect cube.

I found out that p is of the form $18n+1$ and $p=19$ is a solution but I am not getting anything further.

$p^2-p-(m^3-1)=0$

$1+4(m^3-1)=k^2$

$4m^3-3=k^2$

every square is $0,1,4 or 7(\mod9)$ and every cube is $0,1 or 8\mod9$

Using this fact we can conclude that $m^3\equiv{1\mod 9}$

Hence $k$ is either $1$ or $-1$ $\mod9$

But $k\equiv{-1\mod9}$ is not possible since $p$ is a prime. So putting $k\equiv{1\mod9}$ in the quadratic formula we get $p\equiv{1\mod9}$ and since $p$ is odd(as $p=2$ is not a solution), $p\equiv{1\mod18}$

@Travis I made quadratic in terms of $p$ and made discriminent perfevt square. Then I took mod(9) Since every cube is either $0,1$ or $-1$ mod(9). — Satvik Mashkaria, Jan 28 '15 at 07:05
You might like to edit your question to include that information. — Travis Willse, Jan 28 '15 at 07:06
Not sure if this helps, but note that if $n = p^2 - p + 1$ then $n-p$ is a perfect square. So you need a cube that is a prime larger than a perfect square. — Tobias Kildetoft, Jan 28 '15 at 07:59
A similar question without the condition that $p$ is prime: $x^2+x+1$ is the cube of a prime. — Martin Sleziak, May 07 '18 at 10:02

score 1 · Answer 1 · edited May 07 '18 at 10:00

1

this is an easier answer: $$$$we have$$p^2-p=p(p-1)=m^3-1=(m-1)(m^2+m+1)$$we can see $p>m-1$ so we have $$p-1|m-1\to p=1\, MODm-1$$and we have$$m^2+m+1=3\, MODm-1$$we know $$p|m^2+m+1$$ so we have$$m^2+m+1=3p\, or\, m^2+m+1\ge (m+2)p$$if $$m^2+m+1=3p$$ we can see p=19,we have $3(m-1)=p-1$ so we have $m^2+m+1=9m-6$ so m=1 or ,m=7,p=19,$$$$if $$m^2+m+1\ge (m+2)p$$ we have$$(m+2)(m-1)\ge p-1$$ so $$m^2+m-1\ge p=(m^2+m+1)/m+2$$because $m>1$ it is clearly impossible.

edited May 07 '18 at 10:00

Martin Sleziak

56,060

answered Jan 29 '15 at 18:31

ali

2,285

ali, this is all one sentence, without any line breaks. It's impossible to read, even if it's correct. – TonyK Jan 29 '15 at 18:48
i edited the answer@TonyK – ali Jan 30 '15 at 05:53
That's much better! – TonyK Jan 30 '15 at 07:43

score 0 · Answer 2 · answered Jan 28 '15 at 10:52

you can work in $Z[1/2+\sqrt{-3}/2]$.we have unique factorization in this ring,and units are $1,-1,1/2+\sqrt{-3}/2,1/2-\sqrt{-3}/2$,and 2 is prime,yo can easily check this facts by the norm $N(a+b\sqrt{-3})=a^2+3b^2$ we have $4m^3=k^2+3=(k-\sqrt{-3})(k+\sqrt{-3})$,we can see $gcd((k-\sqrt{-3}),(k+\sqrt{-3})|2\sqrt{-3}$ and because$(k,3)=1$ we have $gcd((k-\sqrt{-3}),(k+\sqrt{-3})=2 or 1 $and we have$4(a/2+\sqrt{-3}b/2)\not=k+\sqrt{-3}$ so we have this cases: $$1:k+\sqrt{-3}=2(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=2(a/2-\sqrt{-3}b/2)^3$$ $$2:k+\sqrt{-3}=(1+\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1-\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$ $$3:k+\sqrt{-3}=(1-\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1+\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$ in case 1 by take imaginary part we have:$2(-3b^3/8+3a^2b/8)=1$ so case 1 is impossible. in case 2 by take real and imaginary part we have : $$-3b^3/8+3a^2b/8=1-k=2p$$ $$a^3/8-9ab^2/8=k+3$$ because $p\not=3$ case 2 is impossible case 3:we have $$-3b^3/8+3a^2b/8=1+k=2-2p$$ $$a^3/8-9ab^2/8=k-3=-2-2p$$ $$-3b^3/8+3a^2b/8+a^3/8-9ab^2/8=-4p$$ i cant go further in this case

This is a problem of INMO, a high school level examination. so there should be a simpler solution other than using Ring Theory. — Satvik Mashkaria, Jan 28 '15 at 15:23

Find all primes $p$ such that $p^2-p+1$ is a perfect cube

2 Answers2