Let $p$ be a prime number. If there's a natural number $k$ so that $k^3+pk^2$ is a perfect cube, prove that $3\mid p-1$.
I found this problem in one of my school's selection contests last year. I thought it looked interesting so I gave it a go. I've yet to make any significant progress. I know that $p \equiv 1\ (\textrm{mod}\ 3)$ and I've tried solving for cases where $k \equiv 0,1,2\ (\textrm{mod}\ 3)$, but this seems impossible so I think I have the wrong idea. I've also tried placing $k^3+pk^2$ between two cubes and solving from there, but this also seems wrong. If anyone can guide me on the right path I'd greatly appreciate it. Thanks in advance.
Best Answer
Suppose $k$ is a positive integer and $p$ is a prime such that $k^3+pk^2$ is a perfect cube.
Consider two cases . . .
Case $(1)$:$\;p{\,\mid\,}k$.
Write $k=p^nj$ with $n\ge 1$ and $\gcd(j,p)=1$.
Then we have $$ k^3+pk^2 = p^{2n+1}j^2(p^{n-1}j+1) $$ hence, noting that the factor $j^2$ is relatively prime to each of the factors $$ p^{2n+1},\;\;\;p^{n-1}j+1 $$ it follows that $j^2$ must be a perfect cube.
Since $j^2$ is a perfect cube, so is $j$.
Claim $n\equiv 1\;(\text{mod}\;3)$.
We can assume $n > 1$, since for $n=1$, the claim is automatic.
Then in the factorization $$ k^3+pk^2 = p^{2n+1}j^2(p^{n-1}j+1) $$ the factor $p^{2n+1}$ is relatively prime to each of the factors $$ j^2,\;\;\;p^{n-1}j+1 $$ hence $p^{2n+1}$ must be a perfect cube, so $n\equiv 1\;(\text{mod}\;3)$, as claimed.
It follows that $p^{n-1}j$ is a perfect cube, equal to $x^3$ say, for some $x\ge 1$.
But then $$ p^{n-1}j+1=x^3+1 $$ is strictly between $x^3$ and $(x+1)^3$, so can't be a perfect cube.
Thus case $(1)$ is impossible.
Case $(2)$:$\;p{\,\not\mid\,}k$.
Then we have $$ k^3+pk^2 = k^2(k+p) $$ hence, noting that the factors $$ k^2,\;\;\;k+p $$ are relatively prime, it follows that they must both be perfect cubes.
Since $k^2$ is a perfect cube, so is $k$.
Thus $k=x^3$ and $k+p=y^3$ for some $x,y\ge 1$, so we get $$ p=y^3-x^3=(y-x)(y^2+xy+x^2) $$ hence since $p$ is prime, we must have $y-x=1$.
Then we get $$ p=y^3-x^3=(x+1)^3-x^3=3x^2+3x+1 $$ hence $p\equiv 1\;(\text{mod}\;3)$, as was to be shown.