= Solution
Assume $Q>1$, so $T(k)<0$ for every <wavenumber>. A <Turing instability> therefore occurs exactly when $D(k)<0$ for some $k$. Put $z=k^2\geq0$. The upward quadratic $1+(1-P)z+Pz^2$ has a negative minimum on this half-line precisely when
$$
P>1,\qquad(P-1)^2>4P.
$$
Combining these gives
$$
\boxed{P>3+2\sqrt2.}
$$
The unstable band is $z_-<k^2<z_+$, where
$$
z_\pm=\frac{P-1\pm\sqrt{(P-1)^2-4P}}{2P}.
$$
The degradation-limited <diffusion> distances are proportional to $\ell_u=\sqrt{D_u/\beta_u}$ and $\ell_v=\sqrt{D_v/\beta_v}$. Thus $P=(\ell_v/\ell_u)^2$ and the criterion says \b[the <inhibitor> must travel more than $(1+\sqrt2)$ times as far as the <activator> before degradation]. Short-range activation and longer-range inhibition permit a local peak to reinforce itself. On a finite domain the allowed wavenumbers must also intersect the unstable band; the displayed condition assumes continuously available modes.
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