Is there $m,n \in \mathbb Z$ satisfies $m \arctan \frac{1}{2}+n \arctan \frac{1}{5}=\frac{\pi}{4}$?

Question

I know that $\arctan \frac{1}{2}+\arctan \frac{1}{3}=\frac{\pi}{4}$ and $2\arctan \frac{1}{2}-\arctan \frac{1}{7}=\frac{\pi}{4}$.

Is there $m,n \in \mathbb Z$ satisfies $m \arctan \frac{1}{2}+n \arctan \frac{1}{5}=\frac{\pi}{4}$?

Answer 1

Størmer in 1899 (see also Wikipedia's section on this) showed that there are only four two-term Machin-like formulas with unit numerators for arctangent arguments:

• the two identities you listed
• the identity listed in comments (Machin's classic one)
• $2\tan^{-1}\frac13+\tan^{-1}\frac17=\frac\pi4$

So $m\tan^{-1}\frac12+n\tan^{-1}\frac15=\frac\pi4$ has no solution in integers, or even rational numbers.