Question

Find the sum of squares of all numbers x such that both expressions (x^2) + 6x and x + (1/x) are
integer numbers.

Answer 1

Let $x^2+6x=m$ and $x+\dfrac{1}{x}=n,$ where $m$ and $n$ are integer numbers.

If $x$ ≠$\ 0$ , then

$\begin{cases} x^2+6x=m\\ x+\dfrac{1}{x}=n \end{cases}<=>\begin{cases} x^2+6x-m=0\\ x^2-nx+1=0 \end{cases}$

$<=>\begin{cases} nx-1+6x-m=0\\ x^2-nx+1=0 \end{cases}$

$(n+6)x=m+1$

If $n=-6$ , then $m=-1$ and system of equations $(1)$ is equivalent to $x^2+6x+1=0$ .

$$x_{1,2}=\dfrac{-6\pm \sqrt{36-4}}{2}=-3\pm \sqrt{8},$$

and $(x_1)^2+(x_2)^2=9+12\sqrt{2}+8+9-12\sqrt{2}+8=34$

If $n$ ≠ $-6$ , then $x=\dfrac{m+1}{n+6}$ is a rational number.

$x$ is one of the roots of $x^2-nx+1=0$ . The roots are $x_{1,2}=\dfrac{n\pm\sqrt{n^2-4}}{2}.$ 

We know, that $x$ is rational, therefore $\sqrt{n^2-4}$ is integer, because square root of integer number is either integer or irrational. If it was irrational, then fraction $\dfrac{n\pm\sqrt{n^2-4}}{2}$ wouldn’t be a rational number.

So, $n^2-4=k^2$ , where $k$ is integer.

$n^2-k^2=4\quad \Leftrightarrow \quad (n-k)(n+k)=4$

$n-k$ and $n+k$ are both even numbers

(if they are odd, then their product is odd; if one number is odd and another one is even, then their sum is odd, but it is equal to $2n$ )

We have two cases:

1) $\begin{cases} n-k=2 \\ n+k=2 \end{cases}$ and then $n=2$, $k=0$ .

2) $\begin{cases} n-k=-2 \\ n+k=-2 \end{cases}$ and then $n=-2$, $k=0.$

If $n=2$ : $x^2-2x+1=(x-1)^2=0$ and $x=1$ . If this case $x^2+6x$ and $x+\dfrac{1}{x}$ are integers.

If $n=-2$ : $x^2+2x+1=(x+1)^2=0$ and $x=-1$ . If this case $x^2+6x$ and $x+\dfrac{1}{x}$ are integers.

So, the sum of squares is $34+1+1=36$