Solving $4\arctan\frac1q-\arctan\frac1p=\frac\pi4$ for integer $p$ and $q$

Question

Solve for $p$ and $q$ integers $$4\arctan\frac1q-\arctan\frac1p=\frac\pi4$$

Can you help me with this equation? I can't find a way to pose it, much less solve it.

Answer 1

$\tan (a-b) = \frac {\tan a - \tan b}{1 + \tan a\tan b}$

$\tan (4\arctan \frac 1q - \arctan \frac 1p) = \tan \frac {\pi}{4}$

$\frac {\tan 4\arctan \frac 1q - \frac 1p}{1+ \tan 4\arctan 1q (\frac 1p)} = 1$

$\tan 4\arctan \frac 1q - \frac 1p = 1+ \tan 4\arctan 1q (\frac 1p)\\ p\tan 4\arctan \frac 1q - 1 = p + \tan 4\arctan 1q\\ \tan 4\arctan \frac 1q = \frac {p+1}{p-1}$

$\tan 4a = \frac {4\tan a - 4\tan^3 a}{1 - 6\tan^2 a + \tan^4 a}$

$\tan 4\arctan \frac 1q = \frac {4\frac {1}{q} - 4\frac {1}{q^3}}{1 - 6\frac {1}{q^2} + \frac {1}{q^4}}\\ \frac {4q^3 - 4q}{q^4 - 6q^2 + 1} = \frac {p+1}{p-1}$

$(-1,0)$ gives an integer solution for $p,q.$

But $\frac {1}{q}$ would then be undefined.

For all integers $\frac 12 \le |\frac {p+1}{p-1}| \le \frac 32$

When $|q|$ is large enough ($|q| > 8$) $|\frac {4q^3 - 4q}{q^4 - 6q^2 + 1}| < \frac 12.$

There are finitely many values to check. There may be an even better way to limit this, but I will leave that up to you.

It does not appear that there are any solutions.