A fraction problem

Question

$$a = x + \frac1x \\b =y + \frac1y \\ c = xy +\frac1{xy} $$

Express $c$ in terms of $a$ and $b$

Answer 1

Multiply all things by the LCD to remove fractions.

$$ax=x^2+1$$

$$by=y^2+1$$

$$cxy=x^2y^2+1$$

Multiply the first two equations to get

$$abxy=x^2y^2+1+x^2+y^2=c+x^2+y^2$$

Adding the first two equations, we get

$$ax+by=x^2+y^2+2$$

Combining these two lines,

$$abxy=c+ax+by-2$$

At this point, I'd just go back and solve for $x,y$ using the quadratic formula,

$$c+a\left(\frac{a\pm\sqrt{a^2-4}}2\right)+b\left(\frac{b\pm\sqrt{b^2-4}}2\right)-2=ab\left(\frac{a\pm\sqrt{a^2-4}}2\right)\left(\frac{b\pm\sqrt{b^2-4}}2\right)$$

Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.

Answer 2

Note that $c$ is not uniquely specified by $a$ and $b$, since solving $a=x+\frac{1}{x}$ for $x$ yields two solutions which are reciprocals of each other, and applying $x\mapsto\frac{1}{x}$ or $y\mapsto\frac{1}{y}$ to $xy+\frac{1}{xy}$ yields $\frac{x}{y}+\frac{y}{x}$. However, applying either substitution again gives back $xy+\frac{1}{xy}$, so $c$ can take on either of those values.

Let $c_1=xy+\frac{1}{xy}$ and $c_2=\frac{x}{y}+\frac{y}{x}$. One can check that $$c_1+c_2=ab$$ and $$c_1c_2=a^2+b^2-4.$$ Use the quadratic formula to solve for the possible values of $c$.

Answer 3

$a=x+1/x\iff (x^2-a x+1=0\land x\ne 0)\iff 0\ne x=(a\pm \sqrt {a^2-1})/2.$ Similarly $0\ne y=(b\pm \sqrt {b^2-1})/2.$ Now compute all possible values of $x y +1/x y. $ There are at most 2 values.