Let $X$ be a $n$-dimensional ($n>2$) smooth projective variety over $k=\bar k$ of positive characteristic. Take a divisor $D\in Pic(X).$ Suppose we know that $\frac{[D]^2}{2}\in H^4(X,\mathbb Z_l(2))$ is algebraic (i.e. is the cohomology class of an element of $\mathrm{CH}^2(X)\otimes\mathbb Z_l$) for two relatively prime $l,l'$ (each being distinct from $char(k)$).
It must be obvious but I cannot find a convincing proof of the following statement: there is a $Z\in \mathrm{CH}^2(X)$ such that $[Z]= \frac{[D]^2}{2}\in H^4(X,\mathbb Z_l(2))\ \forall l\neq char(k).$
Edit: $X$ is an abelian variety so that $[D]^2/2$ does make sense ($H^4 = \wedge^4 H^1$) and $l=2$ and $l'$ odd.