Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and  $C$  are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives  $ p \le 3 - 2\sqrt2 $. Or  $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B  = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know   $ \cos (B+C)  =  1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$.  $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives 
$ P \lt 0 $  or $ p \ge 3+ 2\sqrt2 $

Let $\tan\beta=t$.
Thus, $t\neq1$, otherwise $\gamma=\frac{\pi}{2},$ which is impossible.
Also, $$0<\beta<\frac{3\pi}{4},$$ which is
$$0<\beta<\frac{\pi}{4}$$ or 
$$\frac{\pi}{4}<\beta<\frac{\pi}{2}$$ or $$\frac{\pi}{2}<\beta<\frac{3\pi}{4},$$ which is
$$0<t<1$$ or
$$t>1$$ or
$$t<-1.$$
Consider three cases.


*

*$t>1$.


By AM-GM we obtain:
$$p=\tan\beta\tan\left(\frac{3\pi}{4}-\beta\right)=t\cdot\frac{-1-t}{1-t}=\frac{t^2+t}{t-1}=\frac{t^2+t-2+2}{t-1}=$$
$$=t+2+\frac{2}{t-1}=3+t-1+\frac{2}{t-1}\geq3+2\sqrt{(t-1)\cdot\frac{2}{t-1}}=3+2\sqrt2.$$
The equality occurs for $t-1=\frac{2}{t-1}$ and since $\lim\limits_{t\rightarrow1^+}p=+\infty$ and $p$ is a continuous function on $(1,+\infty)$, we  got a range of $p$ in this case: $[3+2\sqrt2,+\infty).$


*$0<t<1$
We obtain:
$$p=-\frac{t(1+t)}{1-t}<0.$$ Also, $$\lim\limits_{t\rightarrow1^-}p=-\infty,$$
$$\lim\limits_{t\rightarrow0^+}p=0$$ and $p$ is a continuous function on $(0,1),$
which says that the range of $p$ in this case it's $(-\infty,0).$ 

*$t<-1$.
In this case $p<0$,  $$\lim_{t\rightarrow-1^-}p=0,$$ $$\lim_{t\rightarrow-\infty}p=-\infty$$ and $p$ is a continuous function on $(-\infty,-1).$
Thus, the range of $p$ in this case it's $(-\infty,0),$ which gives the answer:
$$(-\infty,0)\cup[3+2\sqrt2,+\infty).$$
