When building $kU/2$ via its Postnikov tower, there are some interesting Massey products that show up in the Steenrod algebra, and I'd like to understand them. I bet these appear somewhere in the litterature, but I have not been able to find a reference. I will call $X = kU/2$, $H = H\mathbb{F}_2$ and I'll use $X_{\leq n}$ for the truncated guy, with homotopy in $\pi_0, \ldots, \pi_n$.

Start with $X_{\leq 0} = H$. The $k_0$-invariant to attach $\pi_2$ is the map $X_{\leq 0} =H \stackrel{Q_1}{\to} \Sigma^3 H$. We take its homotopy fiber to get the stage $X_{\leq 2}$.

Now we can rotate and we have $\Sigma^{-1} X_{\leq 0} \stackrel{Q_1}{\to} \Sigma^2 H \to X_{\leq 2}$. The $k_1$-invariant should be given by the extension of $\Sigma^2 H \stackrel{Q_1}{\to} \Sigma^5 H$ to $X_{\leq 2}$. Since $Q_1 Q_1 \simeq 0$ this map does extend to the Postnikov truncation $X_{\leq 2} \stackrel{k_1}{\to} \Sigma^5 H$ giving us the $k$-invariant. (here I am not sure about uniqueness, maybe it is easier to put an $HZ$ at the bottom $\pi_0$, because then $k_1$ would be unique as $[HZ, H] = A/Q_0$).

Anyways, we take the fiber again and get $X_{\leq 4}$. The $k_2$-invariant here will exists if and only if $Q_1 k_1 \simeq 0$. By drawing the diagram defining the Massey product $\langle Q_1, Q_1, Q_1 \rangle$ we see that the product $Q_1 k_1 = 0$ if and only if the product $\langle Q_1, Q_1, Q_1 \rangle$ is $Q_1$-divisible.

How to show this ? I am of course not interested in arguing that $kU$ exists and so this has to happen. Thank you for any help, or any reference.