Any more of this $a\arctan\left(\frac{1}{\phi^n}\right)+b\arctan\left(\frac{1}{\phi^m}\right)=\frac{\pi}{4}$ type?

Asked May 18 '16 at 08:10

Active May 18 '16 at 08:10

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$$\arctan\left(\frac{1}{\phi}\right)+\arctan\left(\frac{1}{\phi^3}\right)=\frac{\pi}{4}$$

$$2\arctan\left(\frac{1}{\phi^2}\right)+\arctan\left(\frac{1}{\phi^6}\right)=\frac{\pi}{4}$$

$$3\arctan\left(\frac{1}{\phi^3}\right)+\arctan\left(\frac{1}{\phi^5}\right)=\frac{\pi}{4}$$

Is there more of this type that yeild $\frac{\pi}{4}$ or these are the only answers?

Where $a,b,m,n \ge 1$ are all integers

$$a\arctan\left(\frac{1}{\phi^n}\right)+b\arctan\left(\frac{1}{\phi^m}\right)=\frac{\pi}{4}$$

I look through many sites, but couldn't find any. If you do come across please let me know. Please no square root involved.

asked May 18 '16 at 08:10

@Arthur Presumably yes, since the three equations hold for the golden ratio. – almagest May 18 '16 at 08:15
@almagest Based on some numerical attempts, I would be more pessimistic than you... – Jean Marie May 18 '16 at 09:09
@JeanMarie Arthur's question has now been deleted. He asked if $\phi$ was the golden ratio. I have no opinion on whether there are other relations of the same type! – almagest May 18 '16 at 09:11
Not quite what you want, but $${\pi\over4}=\sum_{n=1}^{\infty}\arctan(F_{2n+1}^{-1})$$ where $F_n$ are the Fibonacci numbers. – Gerry Myerson May 18 '16 at 09:14
Thank you for your effort @GerryMyerson – May 18 '16 at 10:34
I checked $0<a,b,m,n<20$ and there are no others except those three. – Tito Piezas III Dec 12 '17 at 14:02

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