Prove that exist a $a_n<0$ by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ with the condition $\Delta:= P^2-4Q < 0$ Suppose $P, Q > 0$ such that $\Delta:= P^2-4Q < 0$. The sequence $\{a_n\}$ is defined by the recursion $a_{n+2}-Pa_{n+1}+Qa_n = 0$ where both $a_1$ and $a_2$ are real and at least one is non-zero.
Then, which is needed to be proved, there exists n > 0 such that $a_n < 0$.

As user1952009
said,
if
$a_{n+2}-Pa_{n+1}+Qa_n = 0
$,
by assuming that
$a_n
=r^n
$,
we get,
after dividing by
$r^n$,
$0
=r^2-Pr+Q
$
or
$r
=\frac12(P \pm \sqrt{D})
$
where
$D
= P^2-4Q
< 0
$.
By considering the two possible
values of $r$,
which I will call
$u$ and $v$,
we can find
$A$ and $B$ such that
$A+B = a_0$
and
$Au+Bv = a_1$.
By induction,
this will allow us to show that
$Au^n+Bv^n
= a_n
$
for all $n$.
For an explicit formula for $A$ and $B$,
we have
$B = a_0-A$,
so
$a_1
=Au+(a_0-A)v
=A(u-v)+a_0v
$
or
$A
=\dfrac{a_1-a_0v}{u-v}
$
and
$B
=a_0-A
=a_0-\dfrac{a_1-a_0v}{u-v}
=\dfrac{a_0(u-v)-a_1+a_0v}{u-v}
=\dfrac{a_0u-a_1}{u-v}
$.
Note that
$u-v
=i\sqrt{-D}
$.
Now consider
$u
=\frac12(P + \sqrt{D})
=\frac12(P + i\sqrt{-D})
$.
$|u|
=\frac14\sqrt{P^2+(-D)}
=\frac14\sqrt{P^2-P^2+4Q}
=\frac12\sqrt{Q}
$.
If
$\dfrac{u}{|u|}
=c
$,
then
$|c| = 1$
and $c$ is in the first quadrant,
so
$c = e^{it}
$
where 
$0 < t < \pi/2$.
Therefore
$u^n
=|u|^n c^n
=|u|^n e^{int}
$
will take values
in all four quadrants.
Similarly,
$v 
=\frac12(P - \sqrt{D})
=\frac12(P - i\sqrt{-D})
=\bar{u}
$,
so that
$v^n
=|u|^n e^{-int}
$
and,
since
$A+B = a_0$
and
$\begin{array}\\
A-B
&=\dfrac{a_1-a_0v}{u-v}-\dfrac{a_0u-a_1}{u-v}\\
&=\dfrac{a_1-a_0v-a_0u+a_1}{u-v}\\
&=\dfrac{2a_1-a_0(u+v)}{u-v}\\
&=\dfrac{2a_1-a_0P}{i\sqrt{-D}}\\
\end{array}
$
$\begin{array}\\
Au^n+Bv^n
&=A|u|^ne^{int}+B|u|^n e^{-int}\\
&=|u|^n(Ae^{int}+B e^{-int})\\
&=|u|^n(A(\cos(nt)+i\sin(nt))+B(\cos(nt)-i\sin(nt)))\\
&=|u|^n(\cos(nt)(A+B)+i\sin(nt)(A-B))\\
&=|u|^n(\cos(nt)a_0+i\sin(nt)\dfrac{2a_1-a_0P}{i\sqrt{-D}})\\
&=|u|^n(\cos(nt)a_0+\sin(nt)\dfrac{2a_1-a_0P}{\sqrt{-D}})\\
&=|u|^nR\cos(nt+\theta)\\
\end{array}
$
with $R$ and $\theta$
being gotten by
the usual method of expressing
a linear combination
of
$\sin(z)$
and
$\cos(z)$
as a constant times
$\cos(z+\theta)$
for some $\theta$.
Since
$0 < t < \pi/2$,
$\cos(nt+\theta)$
will take both positive
and negative values
for large enough $n$.
And we are done.
(Whew!)
(added later)
Explicitly,
$\begin{array}\\
R^2
&=a_0^2+\left(\dfrac{2a_1-a_0P}{\sqrt{-D}}\right)^2\\
&=a_0^2+\dfrac{(2a_1-a_0P)^2}{-P^2+4Q}\\
&=\dfrac{a_0^2(-P^2+4Q)+4a_1^2-4a_0a_1P+a_0^2P^2}{-P^2+4Q}\\
&=\dfrac{4a_0^2Q+4a_1^2-4a_0a_1P}{-P^2+4Q}\\
&=4\dfrac{a_0^2Q+a_1^2-4a_0a_1P}{-P^2+4Q}\\
\end{array}
$
and a related expression for
$\tan(\theta)$.
