<s>[INST] Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$ 
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$

I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
 [/INST] A slightly faster variant of the same computation using the identity $$\tan^{-1} u \pm \tan^{-1} v = \tan^{-1} \frac{u \pm v}{1 \mp u v}$$ can be performed by observing that in the special case $u = v$ $$2\tan^{-1} u = \tan^{-1} \frac{2u}{1-u^2}.$$  Consequently, we iterate $g(u) = 2u/(1-u^2)$ twice for $u = 1/5$ to obtain $$4 \tan^{-1} \frac{1}{5} = \tan^{-1} g(g(\tfrac{1}{5})) = \tan^{-1} \frac{120}{119}.$$  Now we apply the original formula to obtain $$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} = \tan^{-1} \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{(119)(239)}} = \tan^{-1} 1 = \frac{\pi}{4}.$$  In all, we used three steps instead of four.
It is also worth noting that when $u, v \in \mathbb Q$, we can write $$\tan^{-1} \frac{p}{q} \pm \tan^{-1} \frac{r}{s} = \tan^{-1} \frac{ps \pm qr}{qs \mp pr}.$$  If we think of each rational as being represented by an ordered pair, which in turn is an element of the complex numbers, e.g. $u = p/q$ has the representation $z = q + pi$, and we define the function $$T(z,w) = \tan\left(\tan^{-1} \frac{\Im(z)}{\Re(z)} + \tan^{-1} \frac{\Im(w)}{\Re(w)}\right),$$ then $$T(z,w) = \frac{\Im(zw)}{\Re(zw)}.$$  In fact, the inverse tangent identity is simply a consequence of multiplication in the complex plane:  $$\arg(zw) = \arg(z) + \arg(w).$$  I leave the details of this relationship as an exercise for the reader.
From the above, we may then regard Machin's formula as a statement about the existence of a nonzero real number $\rho$ such that $$(5+i)^4 = \rho(1+i)(239+i).$$  What is this number?
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