Let $\beta: H^n(X, \mathbb{Z}\_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such that 
\begin{eqnarray\*}\beta \text{Sq}^k=f\beta\end{eqnarray\*}
where $\text{Sq}^k: H^n(X, \mathbb{Z}\_2)\to H^{n+k}(X, \mathbb{Z}\_2)$ is the $k$-th Steenrod square?


Added: It seems that when $k=2$ it is not possible. If we precompose both sides by $\rho\_2:H^n(X, \mathbb{Z})\to H^n(X, \mathbb{Z}\_2)$, the reduction mod 2 map, then the LHS becomes the third differential of the Atiyah-Hirzebruch spectral sequence which in general does not vanish, but the RHS vanishes because $\beta\rho\_2=0$.



An integral cohomology operation related to Steenrod square?
From the Adem relations we get $Sq^1Sq^{k-1}=(k-2)Sq^k$. It follows that for $k$ odd we have $Sq^k=Sq^1Sq^{k-1}$. Note that we have $Sq^1=\rho\beta$, where $\rho$ is mod 2 reduction, so if $k$ is odd $\beta Sq^k=\beta Sq^1Sq^{k-1}=\beta\rho\beta Sq^{k-1}=0$ since $\beta\rho\simeq \ast$. Therefore we are forced to take $f$ such that $f\beta=0$. Working stably we would then get a map $g:H\mathbb{Z}\rightarrow\Sigma^{k}H\mathbb{Z}$ such that $2g=f$. That is $f$ is divisible by 2. Alternatively, any map $H\mathbb{Z}\rightarrow\Sigma^{k}H\mathbb{Z}$ that is divisible by two defines an $f$ with the requried properties. There are many such examples odd-primary torsion classes that work, as well as higher 2-torsion, although it is clear that they do not exist for every odd $k$.


For $k$ even $\beta Sq^k$ is nontrivial since $Sq^1Sq^k=Sq^{k+1}$. Therefore $f\beta=\beta Sq^k$ must be non-trivial as well but this cannot happen. Assume to the contrary that such an $f$ exists. Then we have that $Sq^kSq^1=Sq^k\rho\beta\in H^{k+1}(K(\mathbb{Z}\_2,n-1);\mathbb{Z}\_2)$ and $Sq^1Sq^kSq^1=Sq^{k+1}Sq^1\in H^{k+2}(K(\mathbb{Z}\_2,n-1);\mathbb{Z}\_2)$ are both non-trivial by Cartan. It follows that


$0\neq Sq^{k+1}Sq^1=Sq^1Sq^{k}=\rho\beta Sq^kSq^1=\rho f\beta Sq^1=0$


Since $\beta Sq^1=0$. This is a contradiction and so $f$ cannot exist when $k$ is even.


