Arctangent Equation

From $\tan\left(2\arctan\dfrac15\right)=\dfrac{2\tan\left(\arctan\dfrac15\right)}{1-\tan^2\left(\arctan\dfrac15\right)}=\dfrac{2\left(\dfrac15\right)}{1-\left(\dfrac15\right)^2}=\dfrac25\times\dfrac{25}{24}=\dfrac5{12}$

Thus, $\tan\left(4\arctan\dfrac15\right)=\dfrac{2\tan\left(2\arctan\dfrac15\right)}{1-\tan^2\left(2\arctan\dfrac15\right)}=\dfrac{2\left(\dfrac5{12}\right)}{1-\left(\dfrac5{12}\right)^2}=\dfrac56\times\dfrac{144}{119}=\dfrac{120}{119}$

From Equation

$\begin{aligned}4\arctan\left(\dfrac15\right)-\arctan\left(\dfrac1x\right)=&\dfrac{\pi}4\\4\arctan\left(\dfrac15\right)=&\dfrac{\pi}4+\arctan\left(\dfrac1x\right)\\\tan\left(4\arctan\dfrac15\right)=&\tan\left(\dfrac{\pi}4+\arctan\dfrac1x\right)\\\dfrac{120}{119}=&\dfrac{\tan\dfrac{\pi}4+\tan\left(\arctan\dfrac1x\right)}{1-\tan\dfrac{\pi}4\tan\left(\arctan\dfrac1x\right)}\\\dfrac{120}{119}=&\dfrac{1+\dfrac1x}{1-\dfrac1x}\\120-\dfrac{120}x=&119+\dfrac{119}x\\1=&\dfrac{239}x\\x=&\boxed{239}\end{aligned}$

---

Notes:

$\tan(\arctan{x})=x$

$\tan(2x)=\dfrac{2\tan{x}}{1-\tan^2x}$

$\tan(x+y)=\dfrac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$