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Pontryagin square and $\frac{1}{2}(\mathcal{P}(x) -x^2) =x \cup_1 Sq^1 x$
The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely,
$$
\mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x.
$$
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