XbenX wrote: Is it true, that for all pairs of non-negative integers $a$ and $b$ , the system \begin{align*} \tan{13x} \tan{ay} =& 1 \\ \tan{21x} \tan{by}= & 1 \end{align*}has at least one solution? Wrong. Choose for example $(a,b)=(5,8)$ and system has no solution. (you find that $x$ must be an odd multiple of $\frac{\pi}2$, and so $\tan$ are undefined)

tanxtan(90-x)=1 now easy but tan90=math error

$tan(x+y) = \frac{tan{x}+tan{y}}{1-tan{x}tan{y}}$ $$tan(13x+ay) = \infty $$$$tan(21x+by) = \infty $$.. .

We claim the negative answer. As a counterexample take $a=8,b=13.$ Then for $m,n\in \mathbb Z$ must hold $$13x+ay=\frac \pi 2+\pi m,\quad 21x+by=\frac \pi 2 +\pi n$$implying $y=y(13b-21a)=(13n-21m-4)\pi\implies \tan (ay)=0,$ contradiction.