Solving Cubic Equation: x^3 - kx + (k + 11) = 0

[Taran1]

Hi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.

Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x

Thanks, Taran

 

[Opalg]

Hi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.

Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x

Thanks, Taran

Hi Taran, and welcome to MHB.

If $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \frac{n^3+11}{n-1} = \frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \frac{n^3+11}{n-1}$, as follows: $$ \begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\ n&2&3&4&5&7&13 \\ k&19&19&25&34&59&184 \end{array}.$$ So the only possible values for $k$ are $19,\ 25,\ 34,\ 59,\ 184$ (which all agree with your condition that $k>17$).

 

[MarkFL]

Hi Taran, and welcome to MHB.

If $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \frac{n^3+11}{n-1} = \frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \frac{n^3+11}{n-1}$, as follows: $$ \begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\ n&2&3&4&5&7&13 \\ k&19&19&25&34&59&184 \end{array}.$$ So the only possible values for $k$ are $19,\ 25,\ 34,\ 59,\ 184$ (which all agree with your condition that $k>17$).

Hello, Chris! (Wave)

This question was posted on another site, and I found your reply so insightful, I took the liberty of posting it there, for the benefit of several there trying to solve it. :)

 

[Taran1]

Hi, Thank you so much! This question had my class stumped. That answer makes so much sense. It's been bugging me for a while and I'm very thankful for your help.

Thanks again, Taran