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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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    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? – 2017-01-10
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    Is $n$ an integer? A positive integer? – 2017-01-10
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    @user236182: An integer (as was stated in the journal problem). – 2017-01-10
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    @user236182 yes, $n$ is a positive integer. – 2017-01-10
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    @vidyarthi According to another user it's any integer. – 2017-01-10
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    @quasi And why do you think its unfair? It is not a contest, isnt it? – 2017-01-10
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    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). – 2017-01-10
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    In the future, to be fair, wait until _after_ the deadline (i.e., for the current problems, wait until January 16, 2017). – 2017-01-10
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    @quasi Of course, if I would submit, I would try to cite the question id. By the way, this link should somewhat treat the issue-"http://meta.math.stackexchange.com/questions/5007/should-we-allow-ongoing-journal-questions – 2017-01-10
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    @vidyarthi: Why would you submit at all since you didn't solve it? – 2017-01-10
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    @vidyarthi: And why would you spoil it for other submitters who might come across the solution posted here? – 2017-01-10
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    @quasi have you read the meta link I posted above? – 2017-01-10
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    @vidyarthi: And why not just wait until the deadline is past, and _then_ post the problem. – 2017-01-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51505/discussion-between-vidyarthi-and-quasi). – 2017-01-10
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    @vidyarthi: Yes, I read the meta. It doesn't assert that what you're doing is fine. – 2017-01-10
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    @vidyarthi: And what you're doing -- in my book -- is definitely not cool. – 2017-01-10
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    @vidyarthi: No need to chat -- just consider the fairness. Over and out. – 2017-01-10
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    @quasi what do you mean-in your book? – 2017-01-10
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    @vidyarthi http://idioms.thefreedictionary.com/in+my+book – 2017-01-10
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    @quasi If you arent intersested in answering, please dont answer! There is no point arguing on this issue. Moreover, most of the questions asked in this site and many others are already located in some or the other source-either a journal, a book, or an exam. I am sure of the fact that few, if any of the problems are, original. – 2017-01-10
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    @vidyarthi: Journal problems are fine -- just not ones that are still "open" for submission. – 2017-01-10
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    @quasi As another argument, you can see the names of people published in the previous issue-"an-anduud problem solving group, Mongolia" . This clearly shows group work and discussion is accepted by the Journal authorities. – 2017-01-10
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    @vidyarthi -- not at all the same. That group qualifies as an individual. Their work was private within that group, not shared on a public forum. – 2017-01-10
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    @quasi ok, suppose I do not mention that the problem was from such a journal, then do you think it is fine? – 2017-01-10
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    @vidyarthi: No, that would be even worse. You definitely should attribute the problem to the journal where it came from, and also, you should wait until after the deadline to post it to MSE. – 2017-01-10
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    Journal questions shouldn't be asked directly as some of these questions keep repeating. – 2018-04-05

2 Answers 2

11

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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    how did you manage to factorize $(k^3-4)^2+11$? – 2017-01-10
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    Bravo. How did you see the factorization? ...BTW :For $k\geq 3$ we have $k^4-k^3-2k^2-3k+9\geq 2k^3-2k^2-3k+9\geq 4k^2-3k+9>9.$ – 2017-01-10
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    how is $k^4-k^3-2k^2-3k+9\gt1$? – 2017-01-10
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    @user254665 You can see [WolframAlpha (link)](http://www.wolframalpha.com/input/?i=(k%5E3-4)%5E2%2B11). 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. – 2017-01-10
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    @user236182 Though undoubtedly great, but I am slightly more interested in solving without using WolframAlpha? Any ideas? – 2017-01-10
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    @user236182 how about sophie-germain identity or its variant? – 2017-01-10
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    For the Q, $k\geq 2$ as $5\leq n+4=k^3.$ I just wanted to point out a very elementary method for the polynomial being $>1$ when $k\geq 3$ . – 2017-01-10
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    @user254665 But the question here doesn't claim $n$ is a positive integer. It's unclear what the reader should assume, but $n$ being an integer is the most general and still correct. – 2017-01-10
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    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." – 2017-01-10
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    @vidyarthi I've edited the answer. – 2017-01-11
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    The latter part can be improved. The factorization is great – 2017-01-12
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    @S.Y What do you not like about the latter part? – 2017-01-12
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    @user236182 just feel you could show $n^2+11 > k^2+k+3$ much easily. – 2017-01-12
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    @user236182 For example, $n^2+11=(k^3-4)^2 + 11 > 3k^2+11$ as $|k^3-4| > k^2$ and $|k^3-4|\ge 3$ for any integer $k$. – 2017-01-12
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    @S.Y Thanks. I've edited my answer. Also, $|k^3-4|\ge k^2$ because when $k=2$, then it's an equality. – 2017-11-03
1

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 $k$.