MHB What are the real numbers that give positive integer roots for a cubic equation?

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    2017
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The discussion focuses on finding real numbers k that result in positive integer roots for the cubic equation 5x^3 - 5(k+1)x^2 + (71k-1)x - (66k-1) = 0. Participants are encouraged to solve the problem and share their findings. Members castor28 and Opalg successfully provided correct solutions to the equation. The thread emphasizes the importance of following the community's guidelines for problem-solving. This mathematical inquiry highlights the relationship between cubic equations and their integer roots.
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Hello, MHB Community! (Wave)

anemone has asked me to fill in for her this week. :)

Here is this week's POTW:


Find all real numbers $k$ that give the three roots of the cubic equation $5x^3-5(k+1)x^2+(71k-1)x-(66k-1)=0$ are positive integers.


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!
 
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I would like to give a vote of thanks to MarkFL for standing in for me while I was unavailable to handle my POTW duties. (Handshake) (Smile)

Congratulations to the following members for their correct solution: (Smile)

1. castor28
2. Opalg

Solution from Opalg:

One solution is always $x=1$, because $$5x^3-5(k+1)x^2+(71k-1)x-(66k-1) = (x-1)(5x^2 - 5kx + 66k-1).$$ So we are looking for integer solutions to the equation $5x^2 - 5kx + 66k-1 = 0.$ Notice that the sum of the roots is $k$. So if the roots are integers then $k$ must also be an integer.

The quadratic formula gives the roots as $$\frac{5k \pm\sqrt{25k^2 - 20(66k-1)}}{10}.$$ Thus $25k^2 - 20(66k-1)$ must be a square, say $25k^2 - 1320k + 20 = n^2.$ Complete the square to write that as $(5k - 132)^2 - 17404 = n^2$. Then factorise the difference of two squares to get $$(5k - 132 - n)(5k - 132 + n) = 17404 = 4\cdot 19\cdot 229.$$ So we are looking for a factorisation $17404 = ab$ (with $0<a<b$) such that $$5k - 132 - n = a, \qquad 5k - 132 + n = b.$$ Then $n = \frac12(b-a)$, so that $a$ and $b$ must be either both odd (clearly impossible) or both even. The only possibilities are therefore $17404 = 2\cdot 8702$ and $17404 = 38\cdot 458$.

If we take $a=2$ and $b = 8702$ then $n = 4350$, and the corresponding value of $k$ is given by $5k - 132 - 4350 = 2$, or $5k = 4484$. But that does not give an integer value for $k$.

However, if we take $a=38$ and $b = 458$ then $n = 210$, and the equation for $k$ is $5k - 132 - 210 = 38$, giving $5k = 380$ and therefore $\boxed{k = 76}$. That is the only allowable value for $k$. The original equation then becomes $5x^3 - 385x^2 + 5395x - 5015 = 0$, with solutions $x = 1,\,17,\,59.$
 
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