Non existence of stable vector bundles on $mathbb{P}^4$ with $c_1=0$ and $c_2=1$

Let $mathrm{E}$ be a vector bundle of rank $2$ with $c_1(mathrm{E})=0$ on $mathbf{P}^4$. Then the Chern character of $mathrm{E}$ can be written as $$mathrm{ch}(mathrm{E})=sum_{m=0}^infty frac{2(-1)^m}{(2m)!}c_2(mathrm{E})^m=2-c_2+frac{1}{12}c_2^2+ldots.$$ The degree $4$ part of the Todd class of $mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $mathrm{E}$ is $$chi(mathrm{E})=2chi(mathrm{O}_{mathbf{P}^4})+frac{1}{12}int_{mathbf{P}^4} big(-35c_2 h^2+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$. The case $c_2=3$ was ruled out by Barth and Elencwajg; the latter is known to frequent this site, no doubt he has much more to say.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $mathrm{T}_{mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $mathrm{E}$ on $mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)