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

[MarkFL]

Hello, MHB Community! (Wave)

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

Here is this week's POTW:

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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.

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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]

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:

Spoiler

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 $x = 1,\,17,\,59.$