Question

Find the sum of squares of all numbers x such that both expressions 9x - x^2 and x + (1/x) are integer numbers.

Answer 1

Let $9x-x^2=m$ and $x+\frac{1}{x}=n$ , where $m$ and $n$ are integer numbers.

If $x$ ≠$\ 0$ , then

$\begin{cases} 9x-x^2=m \\ x+\frac{1}{x}=n \end{cases} \quad \Leftrightarrow \quad \begin{cases} x^2-9x+m=0 \\ x^2+1=nx \end{cases} (1) \quad \Leftrightarrow \quad \begin{cases}-9x+m-1=-nx \\ x^2-nx+1=0 \end{cases}$

$x(9-n)=m-1$

If $n=9$ , then $m=1$ and system of equations $(1)$ is equivalent to $x^2-9x+1=0$ .

$x_{1,2}=\frac{9\pm \sqrt{81-4}}{2}=\frac{9\pm\sqrt{77}}{2}$ , and $(x_1)^2+(x_2)^2=\frac{81+18\sqrt{77}+77}{4}+ \frac{81-18\sqrt{77}+77}{4}=79$

If $n$ ≠ $9$ , then $x=\frac{m-1}{9-n}$ is a rational number.

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

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 $\frac{n\pm\sqrt{n^2-4}}{2}$ 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 $9x-x^2$ and $x+\frac{1}{x}$ are integers.

If $n=-2$ : $x^2+2x+1=(x+1)^2=0$ and $x=-1$ . If this case $9x-x^2$ and $x+\frac{1}{x}$ are integers.

So, the sum of squares is $79+1+1=81$

Answer: the sum of squares of all numbers x such that both expressions $9x-x^2$ and $x+\frac{1}{x}$ are integer numbers is equal to 81.