For F-distribution, find df1 and df2 given mode of PDF and height of PDF at mode? Are there closed form equations to find the degrees of freedom of an F-distribution (df1 and df2), given the mode of the PDF and the height of the PDF at the mode? We can assume df1 > 2. If the equations exist, what are they?
 A: The mode of the $F_{\nu_1,\nu_2}$ is at $\frac{\nu_1-2}{\nu_1}\!\cdot\!\frac{\nu_2}{\nu_2+2}$ (for $\nu_1>2$).
The density is 
$$\frac{1}{\mathrm{B}\!\left(\frac{\nu_1}{2},\frac{\nu_2}{2}\right)} \left(\frac{\nu_1}{\nu_2}\right)^{\frac{\nu_1}{2}} x^{\frac{\nu_1}{2} - 1} \left(1+\frac{\nu_1}{\nu_2}\,x\right)^{-\frac{\nu_1+\nu_2}{2}}$$
(e.g. see Wikipedia on the F-distribution)
If you know the mode (say $m$) and the density at the mode (say $\dot{f}$), then solving 
\begin{eqnarray}
m&=&\frac{\nu_1-2}{\nu_1}\!\cdot\!\frac{\nu_2}{\nu_2+2}\\
\dot{f}&=&\frac{1}{\mathrm{B}\!\left(\frac{\nu_1}{2},\frac{\nu_2}{2}\right)} \left(\frac{\nu_1}{\nu_2}\right)^{\frac{\nu_1}{2}} m^{\frac{\nu_1}{2} - 1} \left(1+\frac{\nu_1}{\nu_2}\,m\right)^{-\frac{\nu_1+\nu_2}{2}}
\end{eqnarray}
simultaneously for ${\nu_1}, {\nu_2}$ will not generally be possible algebraically; you would have to proceed numerically as whuber explains in comments.
Note that it's easy to rearrange the first equation to get $\nu_1$ in terms of $m$ and $\nu_2$ (or to get $\nu_2$ in terms of $m$ and $\nu_1$), so that you can eliminate one of the d.f. variables and write the second equation in terms of $\dot{f}$, $m$ and one of the two d.f. variables. 
Since you then have an equation in only one unknown, identifying the numerical solutions is not so onerous.
As an example, consider $m=0.4$ and $\dot{f}=0.65$. Eliminating $\nu_1$ as described, we can plot $\dot{f}$ as a function of $\nu_2$:

and readily identify an approximate solution to $\dot{f}=0.65$, yielding $\nu_2\approx 5.7$ (indeed, I calculated f(5.6) and f(5.7) in order to draw the plot, which were the two points with calculated function values closest to 0.65); a step or two of a suitable root-finding algorithm (regula falsi is fine in this case, though many stats packages will provide some more stable form of root-finding) quickly gives the result $\nu_2\approx 5.6683$.
We can then back out $\nu_1$, and get $\nu_1\approx 4.3586$.
A quick calculation of the original equations confirms these give the correct $m$ and $\dot{f}$ (to similar accuracy as we calculated $\nu_1$ and $\nu_2$ to).  
