If $n^2+11$ is a prime then $n+4$ is not a cube

Question

If $n^2+11$ is a prime number, then is the number $n+4$ a perfect cube?

I think no, but am unable to prove it. By reducing $n^2+11$ modulo $3$ and $10$, and using Fermat's little theorem, I am able to conclude that $n=6k$ or $n=30k$ for some integer $k$. But now, I am unable to proceed further. Any help.

PS: This is problem $S-393$ of Mathematical Reflections.

Answer 1

Look at the contrapositive. If $n+4=k^3$ for some $n,k\in\mathbb Z$, then

$$n^2+11=\left(k^3-4\right)^2+11=$$

$$=\left(k^2 + k + 3\right)\left(k^4 - k^3 - 2 k^2 - 3 k + 9\right)$$

Edit: $n^2+11>k^2+k+3$ because $(k^3-4)^2+11\ge 3k^2+11>k^2+k+3$ because $|k^3-4|\ge 3$, $|k^3-4|\ge k^2$ for all $k\in\mathbb Z$ because if $k\ge 2$, then $k^3\ge 2k^2\ge k^2+4$, if $k\le 0$, then $k^3\le -k^2<-k^2+4$, if $k=1$, then it's true.

Answer 2

Beginning as in user236182's answer, $n+4=k^3$ implies

$$n^2+11=k^6-8k^3+27$$

This suggests first finding the roots of the quadratic $x^2-8x+27$, namely $x=4\pm\sqrt{-11}$, and then hoping that $4+\sqrt{-11}$ can be identified as a cube of the form $(a+b\sqrt{-11})^3$. Doing so will pinpoint $k^2-2ak+(a^2+11b^2)$ as a factor of $k^6-8k^3+27$.

A quick experiment shows that

$$(1+\sqrt{-11})^3=1+3\sqrt{-11}-33-11\sqrt{-11}=-32-8\sqrt{-11}=-8(4+\sqrt{-11})$$

This implies

$$4+\sqrt{-11}=\left(-1-\sqrt{-11}\over2\right)^3$$

which in turn implies $k^2+k+3$ is a factor of $k^6-8k^3+27$. Patient long division (or peeking at user236182's answer) produces

$$k^6-8k^3+27=(k^2+k+3)(k^4-k^3-2k^2-3k+9)$$

For $P(k)=k^4-k^3-2k^2-3k+9$, note that $P'(k)=4k^3-3k^2-4k-3=(4k-3)(k^2+1)$, so that $P(k)$ has a global minimum at $k=3/4$, from which we conclude

$$P(k)\ge P(3/4)\gt0-1-2-3+9=3\gt1$$

For $k^2+k+3$, we have

$$k^2+k+3=\left(k+{1\over2}\right)^2+{11\over4}\gt1$$

Thus $n^2+11=k^6-8k^3+27$ is the product of two integers, $k^2+k+3$ and $k^4-k^3-2k^2-3k+9$, each of which is greater than $1$ for all values of $k$, hence $n^2+11$ is not a prime.

Remark: The key step here was finding the cube root of $4+\sqrt{-11}$. There is no a priori reason to expect there to be a nice cube root (beyond the fact that $4^2+11=27$ is a cube), so success involves a mixture of experience, intuition, and luck. It's possible to take a more systematic approach to searching for the cube root using some elementary theory about algebraic integers, but in this case it was easier (for me, at least) to guess.

It was also partly a matter of luck that $P'(k)$ factored with just one real root, but this was more convenience than necessity; there are plenty of other ways to show that $P(k)$ is never equal to $\pm1$ for integer values of $k$.