<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] For $A,B,C$ to be the angles of a triangle, not only it shall be $A+B+C = \pi$,
but also $0 \le A,B,C$, or strictly greater than $0$ if you exclude the degenerate case.
So, the correct formulation of the problem is
$$
\left\{ \matrix{
  A = \pi /4 \hfill \cr 
  \tan B\tan C = p \hfill \cr 
  A + B + C = \pi  \hfill \cr 
  0 \le A,B,C \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  \tan B\tan C = p \hfill \cr 
  B + C = 3\pi /4 \hfill \cr 
  0 \le B,C\left( { \le 3\pi /4} \right) \hfill \cr}  \right.
$$
Taking care of this further restiction, whatever approach you follow (correctly) you will get to the
unique right solution.
For instance, given $A=\pi /4$, we may start from the symmetrical case , isosceles triangle
$B=C=3 \pi /8$ and put
$$
\eqalign{
  & \left\{ \matrix{
  B = 3\pi /8 + D \hfill \cr 
  C = 3\pi /8 - D \hfill \cr 
   - 3\pi /8 \le D \le 3\pi /8 \hfill \cr 
  \tan \left( {3\pi /8 + D} \right)\tan \left( {3\pi /8 - D} \right) = p \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  B = 3\pi /8 + D \hfill \cr 
  C = 3\pi /8 - D \hfill \cr 
   - 3\pi /8 \le D \le 3\pi /8 \hfill \cr 
  t = \tan \left( {3\pi /8} \right) = \sqrt 2  + 1 \hfill \cr 
   - t \le x = \tan D \le t \hfill \cr 
  {{t^2  - x^2 } \over {1 - t^2 x^2 }} = p \hfill \cr}  \right.  \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  t^2  = 3 + 2\sqrt 2  \hfill \cr 
  0 \le x^2  \le t^2  \hfill \cr 
  p(x) = {{t^2  - x^2 } \over {1 - t^2 x^2 }} \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad p \in \left[ {t^2  = 3 + 2\sqrt 2 , + \infty } \right) \cup \left( { - \infty ,0} \right] \cr} 
$$
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