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

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The problem involves finding all possible values of \( a \) such that the polynomial \( x^3 + ax + 4 - (2 \times 2016^n) = 0 \) has integer roots for \( n \ge 0 \). A clarification was made regarding the value of \( n \), confirming it should be non-negative. The thread encourages members to submit their solutions, highlighting the importance of accuracy in problem statements. Congratulations were given to a member for providing a correct solution. The discussion emphasizes engagement and clarity in mathematical problem-solving.
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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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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!
 
Congratulations to castor28 for his correct solution! (Smile)

Suggested solution:
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 $a=-3$.
 
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