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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.

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    $\begingroup$ Why do you think it's fair to ask for help on a current problem of Mathematical Reflections when the deadline for submission (Juanuary 15, 2017) is not yet passed? $\endgroup$ – quasi Jan 10 '17 at 8:31
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    $\begingroup$ The submitters get ther names published, so if you submit a solution obtained from MSE, you should credit the MSE member, not yourself. If you choose not to submit a solution, it's still unfair to other submitters, since the solution can be Googled (due to you having introduced it as a question in MSE). $\endgroup$ – quasi Jan 10 '17 at 8:41
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    $\begingroup$ In the future, to be fair, wait until after the deadline (i.e., for the current problems, wait until January 16, 2017). $\endgroup$ – quasi Jan 10 '17 at 8:43
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    $\begingroup$ @vidyarthi: Why would you submit at all since you didn't solve it? $\endgroup$ – quasi Jan 10 '17 at 8:46
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    $\begingroup$ @vidyarthi: Journal problems are fine -- just not ones that are still "open" for submission. $\endgroup$ – quasi Jan 10 '17 at 8:59
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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.

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  • $\begingroup$ how did you manage to factorize $(k^3-4)^2+11$? $\endgroup$ – vidyarthi Jan 10 '17 at 7:58
  • $\begingroup$ Bravo. How did you see the factorization? ...BTW :For $k\geq 3$ we have $k^4-k^3-2k^2-3k+9\geq 2k^3-2k^2-3k+9\geq 4k^2-3k+9>9.$ $\endgroup$ – DanielWainfleet Jan 10 '17 at 8:03
  • $\begingroup$ how is $k^4-k^3-2k^2-3k+9\gt1$? $\endgroup$ – vidyarthi Jan 10 '17 at 8:07
  • $\begingroup$ @user254665 You can see WolframAlpha (link). Also, the claim ($k^4-k^3-2k^2-3k+9>1$) is true for all $k\in\mathbb Z$, though you can prove it more easily for special cases as you've shown. $\endgroup$ – user236182 Jan 10 '17 at 8:08
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    $\begingroup$ The actual problem (from Mathematical Reflections, current issue) is as follows: "If $n$ is an integer such that $n^2 + 11$ is a prime, prove that $n + 4$ is not a perfect cube." $\endgroup$ – quasi Jan 10 '17 at 8:34
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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 $k$.

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