This is an elaboration on a question/answer posted on MathOverflow.
As per the MO question, the actual question is (with the typo corrected)
Suppose the cocycle $u\in C^{2p}(X;Z)$ satisfies $\delta u=2a$ for some $a$.
i. Show that $u \cup_0 u + u\cup_1 \delta u$ is a cocycle mod 4.
ii. Define a natural operation, the Pontrjagin square, $P_2:H^{2p}(-;Z_2)\rightarrow H^{4p}(-;Z_4)$.
iii. Show that $\rho P_2(u)=u\cup u$, where $\rho:H^*(-;Z_4)\rightarrow H^*(-;Z_2)$ denotes reduction mod 2.
iv. Show that $P_2(u+v)=P_2(u)+P_2(v)+u\cup v$, where $u\cup v$ is computed with the non-trivial pairing $Z_2 \otimes Z_2\rightarrow Z_4$.
I am quite stuck on part (iv). If you just plug in, expand everything and simplify, you get
$P_2(u+v)=P_2(u)+P_2(v)+u\cup_0 v+v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u$.
Now the answer given on MathOverflow is that $u \cup_0 v$ is not quite commutative "you need to subtract off a coboundary (involving the cup-1 product). So essentially you need to take the expression that you already have and introduce correction coboundary terms (which don't change the cohomology class) to reduce it to the form you're interested in."
Now the only real cup-1 product that seems viable is $$\delta(u \cup_1 v) = -\delta u \cup_1 v - u \cup_1 \delta v + u\cup_0v - v \cup_0 u$$
(or swap $u$ and $v$ in the formula). I write -1, but I am thinking of the coefficients modulo 4.
No matter what I do I can't seem to get the expression to simplify nicely. Moreover, I realised I can't work out what type of expression gives $u \cup v$ - I don't quite understand the part about computing it using the non-trivial pairing $Z_2 \otimes Z_2\rightarrow Z_4$. What sort of expression should I be seeking?
Update: I'll post some more working here. I am focusing on the expression $$f=u\cup_0 v+v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u$$
From the coboundary formula above we see we can write this as (noting that $\delta(u \cup_1 \delta v)$ does not change the cohomology class, and taking everything modulo 4)
$$ \begin{align} f &= \delta u \cup_1 v + u \cup_1 \delta v + v \cup_0 u + v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u \\ &= \delta u \cup_1 v + 2\left(v \cup_0 u\right) +v\cup_1 \delta u \end{align} $$
Now consider $\delta(\delta u \cup_2 v)$. By the coboundary formula we have (again modulo 4)
$$\delta(\delta u \cup_2 v) = - \delta u \cup_1 v - v \cup_1 \delta u$$ and thus we have $$f = 2\left(v \cup_0 u\right)$$
and so
$$P_2(u+v)=P_2(u)+P_2(v)+2\left(v \cup_0 u\right)$$
Is this right? I'm still interested in how to get to an expression involving $u \cup v$?
