<s>[INST] 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 $
 [/INST] I think that every approach should give correct solution. Different solutions appear if to do mistakes.
Conditions to $\angle B$ and $\angle C$ are WLOG
\begin{cases}
\angle B+\angle C = \dfrac{3\pi}4\\[4pt]
\tan\angle  B\tan\angle  C = p\\[4pt]
\angle B\in\left(0,\dfrac{3\pi}8\right]\\
\angle C\in\left[\dfrac{3\pi}8,\dfrac{3\pi}4\right).\tag1
\end{cases}
Also, are known relations

$$\tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a \tan b},\tag{2a}$$
$$\tan\dfrac34\pi = -1,\tag{2b}$$
$$\tan\dfrac38\pi = \frac{1+\cos\frac34\pi}{\sin\frac34\pi}=\sqrt2+1.\tag{2c}$$

Let $x=\tan \angle B$ and $y=\tan \angle C.$ 
then
$$y = \tan\left(\dfrac{3\pi}4-x\right) = \dfrac{-1-x}{1-x} = \dfrac{x+1}{x-1},\tag3$$
$$xy=p.$$
Therefore,
$$P(p,x) = x^2+x-p(x-1) = 0.\tag4$$
If $\,\underline{x=y=\sqrt2+1}\,$ then 
$$P(p,\sqrt2+1) = 4+3\sqrt2 -p\sqrt2 = 0,\\$$
$$p=2\sqrt2+3\tag5.$$
If $\,\underline{x\in (0, \sqrt2+1)}\,$ then 
$P(p,z)\,$ has not roots in the interval $\,z\in(-1,0)\,$ and has one root in the interval $\,z\in(0,\sqrt2+1),\,$
\begin{cases}
P(p,-1)P(p,0) > 0\\[4pt]
P(p,0)P(p,\sqrt2+1)<0,
\end{cases}
\begin{cases}
(p+1)p > 0\\[4pt]
p(2\sqrt2+3-p)<0,
\end{cases}
$$p\in(-\infty,0)\cup(2\sqrt2+3,\infty).$$
Taking in account $(5),$ the answer is
$$\color{green}{\boxed{{\phantom{\Big|}\mathbf{p\in(-\infty,0)\cup[2\sqrt2+3,\infty)}.}}\tag6}$$
If the condition $(6)$ is satisfied, then the common solution is
$$(\angle B,\angle C)\in\{(f_\angle(z_1),f_\angle(z_2)), (f_\angle(z_2),f_\angle(z_1))\},\tag7$$
where
$$D=p^2-6p+1 = (p-3-2\sqrt2)(p-3+2\sqrt2),\tag8$$
$$z_{1,2} = \dfrac{p-1\pm\sqrt D}2,\tag9$$
$$f_\angle(z) = \arctan z +\dfrac\pi2(1-\operatorname{sgn} z).\tag{10}$$
Example 1. $\quad p=-\frac32,\quad z\in\{-3,\frac12\},\quad \angle B\approx 27^\circ,\quad \angle C \approx 109^\circ,\quad\angle B+\angle C = 135^\circ =\frac34\pi. $
Example 2. $\quad p=6,\quad z\in\{2,3\},\quad \angle B\approx63^\circ,
\angle C \approx 72^\circ,\quad \angle B +\angle C = 135^\circ -\frac34\pi.$
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