Integers as Roots of Polynomial: Find Possible Values | POTW #293 Dec 19th, 2017

[anemone]

Here is this week's POTW:

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Suppose $n \ge 0$ and all the roots of $x^3+ax+4-(2\times 2016^n)=0$ are integers. Find all possible values of $a$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Hi MHB!

I want to apologize for the $n$ for this week problem should be always greater than or equal to zero, rather than it is a positive integer and I have made the respective change to above post.

I am truly sorry for making this typo and I hope this clears some members confusion and I am looking forward to receive more submissions to this week problem!

 

[anemone]

Congratulations to castor28 for his correct solution! (Smile)

Suggested solution:

Spoiler

Let the integer roots of the equation be $m,\,n$ and $k$. Then by Vieta's formulas, we have

$m+n+k=0,\,mn+nk+km=a,\,mnk=2\times 2016^n-4$.

Assume that $n\ge 1$, then $7 \nmid m,\,n,\,k$ and from $(m+n+k)^3=0$ we get

$m^3+n^3+k^3=3mnk\equiv 2(\mod 7)$

Note that the non-zero cubic residues modulo 7 are 1 and 6 and so we cannot have $m^3+n^3+k^3=3mnk\equiv 2(\mod 7)$.

Thus $n=0$, $\therefore m+n+k=0$ and $mnk=-2$, which gives $(m,\,n,\,k)=(1,\,1,\,-2)$ and its cyclic permutations.

Therefore $a=-3$.