Do you know the product?

$\begin{aligned} \frac {a+b}{a-b} + \frac {a-b}{a+b} & = \frac {17}4 \\ \frac {(a+b)^2+(a-b)^2}{(a-b)(a+b)} & = \frac {17}4 \\ \frac {2(a^2+b^2)}{a^2-b^2} & = \frac {17}4 \\ 8a^2+8b^2 & = 17a^2-17b^2 \\ 9a^2 & = 25b^2 \\ \frac {a^2}{b^2} & = \frac {25}9 \\ \frac ab & = \frac 53 \end{aligned}$

Therefore, $a=5$ and $b=3$ and $ab = \boxed{15}$ .

$\begin{aligned} & \dfrac{a+b}{a-b} +\dfrac{a-b}{a+b} = \dfrac{17}{4} \\& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{17}{8} \phantom{ccccc}\\& \dfrac{a^2+b^2}{a^2-b^2} =\dfrac{2}{2}\left(\dfrac{17}{8}\right)= \dfrac{34}{16} \\& a^2+b^2 =34 \phantom{cccccc} a^2-b^2 = 16\end{aligned}$ Solving for $a$ and $b$ we get $a=5$ and $b=3$ . So $a \times b = 15$ .